The Xenharmonic Wiki is a great resource to start understanding and, if enough determination is available, constructing temperaments as well as scales. As some of you may know, this Wiki is not beginner friendly.
With that out of the way, let's get to the question: On the page "Optimal patent val" the author(s) state that
Given any collection of p-limit commas, there is a finite list of p-limit patent vals tempering out the commas. [...] [A]mong these patent vals will be found the unique patent val which has the lowest TE error.
Naturally, I tried to find out how to calculate the Tenney-Euclidean Error with the help of the corresponding page. After some fiddling around, I figured out that the wedge products are calculated with the standard vector basis (?) and this page shows a kind of tongue-in-cheek way to calculate the produt for two "1-vals" (vectors). On a side note, this paper demonstrates a way to calculate the complexity just using vectors and matrices. These are the equations given above:
$\sqrt{\det{[\frac{MW^2M^T}{HW^2H^T}}]}$ for the temperament complexity
$\sqrt{\det{[\frac{MW^2M^T}{HW^2H^T}(1+E^2_k)-\frac{MW^2H^THW^2M^T}{(HW^2H^T)^2}}]}$ for the temperament "badness"
The error is simply the quotient of the complexity and the "badness".
$M$ describes the val that contains the mapping of the generator (usually one step of an EDO) to the $\log_2$ of the primes up to a prime-limit.
$H$ is a row vector containing said $\log_2$ of the primes.
$W$ is a square weighting matrix that, for "Tenney weighting", just contains the reciprocal of the values described for $H$ on its diagonal.
Lastly, $E_k$ is the smallest error that you care about, so it is a free parameter.
For complexity, you just wedge all starting vals together and call the result matrix $M$ (and take the norm if you want to know about it). For badness, it simply is the norm of the wedge product of the matrix $M$ with $J$ (a vector with all entries being 1).
Now, the paper gives a way to calculate "badness" without the wedge product (as you can see above), but how would I go about calculating "complexity" and "badness" without cheesing my way through the calculation? Since wedging this multivector (I presume) with a vector makes... little sense to me. Hence the question.
At last, I know that the wedge (a.k.a. exterior) product is dependent on the used basis, but an intuitive explanation would be appreciated. Thanks.
I'll rephrase the definitions from your links in terms of abstract linear algebra (no matrices).
A monzo (vector/ket) $\vec x$ can be written as
$$\vec x=|x_2\;x_3\;x_5\;x_7\;\cdots\;x_{p_n}\rangle=x_2\vec b_2+x_3\vec b_3+x_5\vec b_5+\cdots$$
which corresponds to the number $2^{x_2}\cdot3^{x_3}\cdot5^{x_5}\cdots$ .
A val (covector/bra) $\vec f$ can be written as
$$\vec f=\langle f_2\;f_3\;f_5\cdots|=f_2\vec\alpha_2+f_3\vec\alpha_3+f_5\vec\alpha_5+\cdots.$$
The sets $\{\vec b_p\}$ and $\{\vec\alpha_p\}$ are dual bases, so that $\vec\alpha_p(\vec b_p)=1$ and otherwise $\vec\alpha_p(\vec b_q)=0$.
The basis monzos are defined to be orthogonal, and to have Euclidean norm $\lVert\vec b_p\rVert_E=\log_2 p$, so they can be made into an orthonormal basis:
$$\hat b_p=\frac{\vec b_p}{\log_2 p};\quad\lVert\hat b_p\rVert_E=1.$$
The Euclidean dot product on monzos induces a dot product on vals, according to which the basis vals are orthogonal, and have norm $\lVert\vec\alpha_p\rVert_E=1/\log_2 p$, so they can also be made into an orthonormal basis:
$$\hat\alpha_p=\vec\alpha_p\log_2 p;\quad\lVert\hat\alpha_p\rVert_E=1,$$
and they're still dual to each other: $\hat\alpha_p(\hat b_p)=1$.
The weighted coordinates of $\vec x$ are the coefficients with respect to the orthonormal basis:
$$\vec x=x_2\vec b_2+x_3\vec b_3+x_5\vec b_5+\cdots=\big(x_2\big)\hat b_2+\big(x_3\log_23\big)\hat b_3+\big(x_5\log_25\big)\hat b_5+\cdots,$$
$$\vec f=f_2\vec\alpha_2+f_3\vec\alpha_3+f_5\vec\alpha_5+\cdots=\big(f_2\big)\hat\alpha_2+\Big(\frac{f_3}{\log_23}\Big)\hat\alpha_3+\Big(\frac{f_5}{\log_25}\Big)\hat\alpha_5+\cdots.$$
The exterior algebra of multimonzos has a basis
$$\{1,\\ \hat b_2,\;\hat b_3,\;\hat b_5,\;\cdots,\\ \hat b_2\wedge\hat b_3,\;\hat b_2\wedge\hat b_5,\;\hat b_3\wedge\hat b_5,\;\cdots,\\ \hat b_2\wedge\hat b_3\wedge\hat b_5,\;\cdots,\\ \vdots\quad\},$$
which also turns out to be orthonormal, according to the dot product defined by the Gramian determinant:
$$\big(\vec x_1\wedge\vec x_2\wedge\cdots\wedge\vec x_r\big)\star\big(\vec y_1\wedge\vec y_2\wedge\cdots\wedge\vec y_r\big)=\det[\vec x_i\cdot\vec y_j].$$
(There's an equivalent definition in terms of geometric algebra; see a recent answer of mine.) This expression is essentially the same as that for applying a multival to a set of monzos:
$$\big(\vec f_1\wedge\vec f_2\wedge\cdots\wedge\vec f_r\big)\big(\vec y_1,\vec y_2,\cdots,\vec y_r\big)=\det[\vec f_i(\vec y_j)].$$
The Euclidean norm of a multimonzo or multival is $\lVert M\rVert_E=\sqrt{M\star M}$. The Tenney-Euclidean norm, or complexity, is
$$\lVert M\rVert_{TE}=\frac{\lVert M\rVert_E}{\sqrt{\binom{n}{r}}}=\sqrt\frac{M\star M}{\left(\frac{n!}{(n-r)!r!}\right)}.$$
The badness of the multival $M$ is defined as $\lVert J\wedge M\rVert_{TE}$, in terms of the val
$$J=\hat\alpha_2+\hat\alpha_3+\hat\alpha_5+\cdots+\hat\alpha_{p_n}.$$
$M$ is not a matrix. If you've been given $M$ as a function (specifying how it acts on monzos), you'll need to express it as a sum of wedge products of vals, in order to calculate its norm.
Example:
From here, 7-limit meantone is supposed to have the syntonic comma $(-4\vec b_2+4\vec b_3-\vec b_5)$ and $(-13\vec b_2+10\vec b_3-\vec b_7)$ in its nullspace. Any linear subspace can be represented by the wedge product of its basis vectors; even if we change the basis, the wedge product is unique up to a scale factor. So this nullspace is represented by
$$N=(-4\vec b_2+4\vec b_3-\vec b_5)\wedge(-13\vec b_2+10\vec b_3-\vec b_7)$$
$$=12\vec b_2\wedge\vec b_3-13\vec b_2\wedge\vec b_5+10\vec b_3\wedge\vec b_5+4\vec b_2\wedge\vec b_7-4\vec b_3\wedge\vec b_7+\vec b_5\wedge\vec b_7.$$
Now, to find the multival $M$ with nullspace $N$... I don't think this operation has a name... "Hodge duality", "Poincare duality", and "volume form" come to mind, but they involve too much structure. Let's call this a volume form anyway: $(\vec\alpha_2\wedge\vec\alpha_3\wedge\vec\alpha_5\wedge\vec\alpha_7)$; it represents the entire val space. We just apply this to $N$ to get $M$. In geometric algebra, this would be a fat dot product:
$$\pm M=(\vec\alpha_2\wedge\vec\alpha_3\wedge\vec\alpha_5\wedge\vec\alpha_7)\bullet N$$
$$=-12\vec\alpha_5\wedge\vec\alpha_7-13\vec\alpha_3\wedge\vec\alpha_7-10\vec\alpha_2\wedge\vec\alpha_7-4\vec\alpha_3\wedge\vec\alpha_5-4\vec\alpha_2\wedge\vec\alpha_5-\vec\alpha_2\wedge\vec\alpha_3$$
$$=-\big(\vec\alpha_2\wedge\vec\alpha_3+4\vec\alpha_2\wedge\vec\alpha_5+4\vec\alpha_3\wedge\vec\alpha_5+10\vec\alpha_2\wedge\vec\alpha_7+13\vec\alpha_3\wedge\vec\alpha_7+12\vec\alpha_5\wedge\vec\alpha_7\big).$$
Compare this with the wiki's "wedgie for septimal meantone". (I disagree with their choice of ordering.) We choose the sign so that the coefficient of $\vec\alpha_2\wedge\vec\alpha_3$ is positive, but that's not important for calculating the norm. I'll use the abbreviation $\lg=\log_2$.
$${\lVert M\rVert_E}^2=M\star M$$
$$=\Big(\frac{1}{\lg3}\Big)^2+\Big(\frac{4}{\lg5}\Big)^2+\Big(\frac{4}{\lg3\lg5}\Big)^2+\Big(\frac{10}{\lg7}\Big)^2+\Big(\frac{13}{\lg3\lg7}\Big)^2+\Big(\frac{12}{\lg5\lg7}\Big)^2$$
$$\lVert M\rVert_E\approx5.4000$$
$$\lVert M\rVert_{TE}\approx2.2046\,.$$
And for your main question:
$$J=\hat\alpha_2+\hat\alpha_3+\hat\alpha_5+\hat\alpha_7$$
$$J\wedge M=\Big(\frac{4}{\lg3\lg5}-\frac{4}{\lg5}+\frac{1}{\lg3}\Big)\hat\alpha_2\wedge\hat\alpha_3\wedge\hat\alpha_5\\+\Big(\frac{13}{\lg3\lg7}-\frac{10}{\lg7}+\frac{1}{\lg3}\Big)\hat\alpha_2\wedge\hat\alpha_3\wedge\hat\alpha_7\\+\Big(\frac{12}{\lg5\lg7}-\frac{10}{\lg7}+\frac{4}{\lg5}\Big)\hat\alpha_2\wedge\hat\alpha_5\wedge\hat\alpha_7\\+\Big(\frac{12}{\lg5\lg7}-\frac{13}{\lg3\lg7}+\frac{4}{\lg3\lg5}\Big)\hat\alpha_3\wedge\hat\alpha_5\wedge\hat\alpha_7$$
$$\lVert J\wedge M\rVert_E\approx0.012435$$
$$\lVert J\wedge M\rVert_{TE}\approx0.0062175\,.$$
...Either I'm wrong, or my calculator's wrong, or the wiki's wrong. It says the badness is $0.013707$ . I tried swapping $\vec\alpha$'s with $\hat\alpha$'s (adding or removing $\lg$'s), but it still doesn't match.
I'll come back later to look for errors.