Why should there be a 7-dimensional cross product in the context of exterior algebra? The three-dimensional cross product can be viewed as the wedge product corresponding to the exterior power $\Lambda^2(\mathbb R^3)$. An explanation that I have come up with for the scarcity of cross products is the following: we have the classical result $$\dim(\Lambda^m(\mathbb R^n))=\binom{n}{m}$$ and in particular $$\dim(\Lambda^2(\mathbb R^3))=\binom 3 2 = 3.$$ Since other $n$ don't  satisfy $\binom n 2 = n$, there will be no other dimensionality in which the wedge product is a function $\mathbb R^n\to\mathbb R^n$ (in particular similar logic applies to why the triple product on $\mathbb R^3$ is a scalar: $\binom 3 3=1$.) In this case, how is my logic losing the possibility of a 7-dimensional cross product, noting that $\binom 7 2 = 21$ and thus the 7d wedge product maps into $\mathbb R^{21}$?
Note: I'm perfectly aware about quaternions and octonions providing an explanation for the existence of a 7d cross product - I'm not asking for a construction.
 A: Although the map $(x,y)\to *(x\wedge y)$ doesn't generalize to a map $\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ in general, not every cross-product has to be generated as such. In fact, the cross-product on $\mathbb{R}^3$ is unique modulo sign, so you could think of the construction you outline as just an effect of that uniqueness: anything that 'looks' like a cross-product (linear, antisymmetric, etc.) has to be the usual one derived from, e.g., the quaternion product.
So, how do we generalize this contruction to higher dimensions? To simplify notation, I'll deal with tensor products rather than wedge products. (The outcome is the same; we just need to remember that we're working with antisymmetric functions throughout). The tensor (or wedge) product gives a map $V\times V \to V^{\otimes 2}$. If we have an element $v\in V^{\otimes 3}$, dualizing gives 
$$v^∗\in\operatorname{Hom}(V^{\otimes 3},\mathbb{R})=Hom(V^{\otimes 2},\operatorname{Hom}(V,\mathbb{R}))=Hom(V^{\otimes 2},V).$$
That's exactly what we're looking for, although we have to check that it's a valid cross-product. For $V = \mathbb{R}^3$, there's a canonical choice of $v$: the element $∗1=e_1 \otimes e_2 \otimes e_3$, which corresponds to a nonzero element of the $1$-dimensional space $\Lambda^3 V$. (Note that we actually care about wedge products here rather than tensor products, despite the sloppy notation). In fact, this construction is exactly what the Hodge star is, just in more generality. In higher dimensions, though, there's no canoncial choice of $v$.
