Cross Product in $\mathbb{R}^3$
In $\mathbb{R}^3$ one can compute the coordinates of the cross product between $b_2$ and $b_3$ using the Levi-Civita symbol as follows: $$b_{1,i_1} = \sum_{i_2i_3} \epsilon_{i_1i_2i_3} b_{2,i_2} b_{3,i_3}.$$
"Cross Product" of $n-1$ Vectors in $\mathbb{R}^n$
I use this to generalise the cross product to $\mathbb{R}^{n}$. That is, given $n-1$ linearly independent vectors $b_2,\ldots,b_n$, the vector $b_{1}$ should be orthogonal to those if its coordinates are defined as $$b_{1,i_{1}} = \sum_{i_2\ldots i_n} \epsilon_{i_1 i_2 \ldots i_n} b_{2,i_2} \ldots b_{n, i_n}.$$ This can be derived for example from the matrix adjugate.
Relation to the Matrix Adjugate
In the case of the $n-1$ vectors $b_2,\ldots,b_n$ in $\mathbb{R}^{n}$, where I compute $b_1$ from them, I could stack all of those together as rows in a $n\times n$ matrix $A$. The adjugate matrix $\operatorname{adj}(A)$ has a first column equal to $b_1$, and we know $A \cdot\operatorname{adj}(A) = \operatorname{det}(A)\cdot I$, thus $b_1$ is orthogonal to $b_2,\ldots,b_n$.
Orthogonal Subspace of $r$ Linearly Independent Vectors in $\mathbb{R}^n$?
More generally, if I have $n-r$ vectors $b_{r+1}, \ldots, b_n$ the contraction with the Levi-Civita symbol yields an $r$-dimensional array: $$B_{i_1,\ldots,i_{r}} = \sum_{i_{r+1}\ldots i_n} \epsilon_{i_1\ldots i_n} b_{r+1,i_{r+1}}\ldots b_{n,i_n}.$$ What is the geometrical meaning of this array? Assuming that $b_{r+1},\ldots,b_n$ are linearly independent, can $B$ be related to the subspace orthogonal to $span(b_{r+1},\ldots,b_n)$?
In general I cannot use the same trick with the matrix adjugate since a number of unknown terms will enter the factors unlike in the case with the $n-1$ given vectors.
$$\begin{bmatrix} ? \\ \vdots \\ ? \\ b^T_{r+1} \\ \vdots \\ b^T_{n} \end{bmatrix}$$
This is consistent with the idea that I have additional freedom on how to choose the basis of the orthogonal subspace. However, I still expect that this is somehow related to this object $B$.
Edit:
@blargoner suggested I try to relate this to the Hodge star, which helped a bit, but I don't think I have the whole picture since I have an issue with coming up with a canonical basis for the orthogonal complement of $span(b_{r+1},\ldots, b_n)$.
Hodge Star
If I have $n-1$ linearly independent vectors $b_2,\ldots,b_n$ then the wedge product of those should yield a $(n-1)$-vector $\tilde{b}_1 = b_2\wedge \ldots \wedge b_{n}$. Using the Hodge star I can convert the $(n-1)$-vector $\tilde{b}_1$ into a vector $b_1 = \star \tilde{b}_1$ with components given as follows $b_1 = (B_{1,2,\ldots,n}, B_{2,3\ldots,n,1}, \ldots, B_{n,1,\ldots,n-1})$. This is all fine as long as I have $n-1$ linearly independent vectors, however things are not as simple when I have $n-r<n-1$ vectors. Then $b_{r+1}\wedge\ldots\wedge b_n$ is an $(n-r)$-vector, and using the Hodge star I can get an $r$-vector $b_1\wedge\ldots\wedge b_r = \star(b_{r+1}\wedge\ldots\wedge b_n)$. However I have some freedom in the choice of the vectors $b_1,\ldots,b_r$ which makes things more complicated.
Basis for the Orthogonal Complement of $span(b_{r+1},\ldots,b_n)$
In the general case, I do not think there is a canonical choice of vectors $b_1,\ldots, b_r$, since I have several degrees of freedom. The equality $b_1\wedge \ldots \wedge b_r = \star(b_{n-r+1}\wedge \ldots \wedge b_n)$ provides the constraints: $$\sum_{i_1\ldots i_{n-r}} \epsilon_{i_1\ldots i_n} b_{1,i_1}\ldots b_{n-r,i_{n-r}} = B_{i_1,\ldots,i_{n-r}} = \sum_{i_{n-r+1}\ldots i_n}\epsilon_{i_1\ldots i_n} b_{n-r+1,i_{n-r+1}}\ldots b_{n,i_n}.$$ This means that the oriented hypervolumes of the projections of the parallelotope $b_1,\ldots, b_r$ on the planes $span(e_{i_{r+1}}, \ldots, e_{i_n})$ must be equal to the oriented hypervolumes of the projections of the parallelotopes $b_{r+1},\ldots,b_n$ on the planes $span(e_{i_1},\ldots,e_{i_r})$.
3D Example
In the $\mathbb{R}^3$ case if I am given two vectors, then the third is fully determined by those as discussed. If I have only a single vector $b_1=(b_{1,1}, b_{1,2}, b_{1,3})$ then for the other two the following constraints must be satisfied: $$b_{2,2}b_{3,3}-b_{2,3}b_{3,2} = b_{1,1}, \quad b_{2,3}b_{3,1}-b_{2,1}b_{3,3} = b_{1,2}, \quad b_{2,1}b_{3,2}-b_{2,2}b_{3,1} = b_{1,3}.$$ There are six unknowns, and the above gives three equations. For simplicity I can imagine that $b_{1,1}=b_{1,2}=0$ and $b_{1,3}\ne 0$, then the vectors $b_2,b_3$ must be in the plane spanned by $e_1,e_2$. As far as I can tell the three constraints guarantee orthogonality, and also prescribe the signed area of the parallelogram spanned by $b_1,b_2$ (in fact $b_1^2+b_2^2+b_3^2$ should give the squared volume of the parallelepiped). I believe that the remaining three degrees of freedom can be taken into account by a rotation of $b_1,b_2$ around $b_3$ (since this preserves signed areas), then length of $b_2$, and the "height" of $b_3$ w.r.t. $b_2$. I am not sure whether there is a canonical way to fix those degrees of freedom.
4D Example
In 4D if 3 linearly independent vectors $b_2,b_3,b_4$ are given, they fully determine a fourth vector orthogonal to those using $b_1 = \star(b_2\wedge b_3 \wedge b_4)$. The more interesting case arises when I have two vectors. Then I get a bivector:
\begin{align}\star(b_1 \wedge b_2) &= \star\left(\left(\sum_{i=1}^4b_{1,i} e_i\right)\wedge \left(\sum_{j=1}^4 b_{2,j} e_j\right)\right) \\ &= \star((b_{1,1}b_{2,2}-b_{1,2}b_{2,1})\,e_1\wedge e_2 +(b_{1,3}b_{2,1} - b_{1,1}b_{2,3})\,e_{3}\wedge e_1 \\ &+(b_{1,1}b_{2,4}-b_{1,4}b_{2,1})\, e_1\wedge e_4 +(b_{1,2}b_{2,3}-b_{1,3}b_{2,2})\, e_2 \wedge e_3 \\ &+(b_{1,4}b_{2,2}-b_{1,2}b_{2,4})\, e_4\wedge e_2 +(b_{1,3}b_{2,4}-b_{1,4}b_{2,3})\, e_3\wedge e_4) \\ &= (b_{1,1}b_{2,2}-b_{1,2}b_{2,1})\,e_3\wedge e_4 +(b_{1,3}b_{2,1} - b_{1,1}b_{2,3})\,e_{2}\wedge e_4 \\ &+(b_{1,1}b_{2,4}-b_{1,4}b_{2,1})\, e_2\wedge e_3 +(b_{1,2}b_{2,3}-b_{1,3}b_{2,2})\, e_1 \wedge e_4 \\ &+(b_{1,4}b_{2,2}-b_{1,2}b_{2,4})\, e_1\wedge e_3 +(b_{1,3}b_{2,4}-b_{1,4}b_{2,3})\, e_1\wedge e_2 \\ &= b_3\wedge b_4. \end{align}
I see no obvious way to assign some permutation of the coordinates of $b_1$ and $b_2$ to $b_3$ and $b_4$ as to get the above wedge. Are there some simple and known canonical ways to assign coordinates to $b_3,b_4$ such that they fulfill the above?