I can do the algebra to prove this identity:
$$(\mathbf{a} \times \mathbf{b}) \cdot ((\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a})) = (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))^2$$
in particular, by appeal to the ``BAC -- CAB'' identity mentioned on Wikipedia plus some simplifications.
Is there a geometric or linear-algebraic interpretation of this? I know that the scalar triple product is the volume of the paralellopiped formed by its components, and it seems this should be related, but I'm not sure how. Where would volume squared come in?
The volume of the parallelepiped is given by $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$. The LHS of your expression is the triple scalar product using the vectors $\mathbf{a}\times\mathbf{b}$, $\mathbf{b}\times\mathbf{c}$ and $\mathbf{c}\times\mathbf{a}$. These vectors represent vectors that are orthogonal to the corresponding faces of the parallelepiped, and they have the magnitude of the area of each parallelogram. So we are looking at a $2n$ dual to the original parallelepiped, with the bounding planes being treated as edges.
To see how we get to 'volume squared', first imagine the parallelepiped is a cube, and we are multiplying the faces. We have $xy.yz.zx=x^2y^2z^2$, thus a volume squared. The same thing applies to the general parallelepiped.