This is an attempt to make this question more specific.
We can compute the length of the cross product in $\mathbb{R}^3$ using the formula: $$|x\times y| = \sqrt{|x|^2|y|^2 - |x\cdot y|^2}$$
This makes sense for $x,y \in \mathbb{R}^3$ and has a geometric meaning; it's the area of the parallelogram spanned by $x$ and $y$.
There's also a higher-dimensional version. For example, if $x,y,z \in \mathbb{R}^4$, there's a natural way of getting an orthogonal vector, that we might as well denote $x \times y \times z$. It should be understood that the function $\Box \times \Box \times \Box$ is a ternary operator, and cannot be refactored as $(\Box \times \Box) \times \Box$ or anything like that.
Anyway, I'd like to know if there's a similar formula for higher-dimensional cross products, perhaps of the form $$|x \times y \times z| = \sqrt{\mathrm{something}}$$
It seems reasonable to conjecture that this is the square root of the determinant of the corresponding Gramian matrix. In particular, it seems reasonable to expect that $$|x \times y \times z| = \sqrt{\det([x,y,z]^\top [x,y,z])}$$
I'm tempted to write this as an answer to the question linked at the beginning of this question, but I'm not 100% sure it's true.
Question. Is the length of the multilinear cross product of $x_1,\ldots,x_n \in \mathbb{R}^{n+1}$ given by $$|x_1 \times \cdots \times x_n| = \sqrt{\det([x_1,\ldots,x_n]^\top [x_1,\ldots,x_n])}\ ?$$
If so, how do we prove this?
If not, what is the correct formula?
I show the proof for $n=3,$ it can easily be extended to any $n\in\mathbb{N}.$
Let $u=x\times y\times z$ and $A=\begin{pmatrix} & & & \\ u & x & y & z \\ & & & \end{pmatrix} \in \mathbb{R}^{4\times 4}.$ Applying Cramer's Rule, we find $$ A^T\cdot u = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \cdot \det(A) $$ On the other hand, we have $u=A\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}.$ Therefore, $$ A^TA\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}\cdot \det(A) $$ Note that $$A^TA = \begin{pmatrix} u^Tu & 0 \\ 0 & \left[x,y,z\right]^T\left[x,y,z\right] \end{pmatrix} $$ Therefore, we get $$ u^Tu = \det(A) = \sqrt{\det(A^T)\det(A)} = \sqrt{\det(A^TA)} =\sqrt{u^T u \cdot \det(\left[x,y,z\right]^T\left[x,y,z\right])} $$ Now we can divide by $\sqrt{u^Tu}$ on both sides and get the desired result.