Let $P=[0,a_1]\times\ldots\times[0,a_n]$ be a parallelepiped in $\mathbb{R}^n$. What is the greatest possible measure of an orthogonal projection of $P$ onto a $d$-dimensional affine subspace in $\mathbb{R}^n$ (identified with $\mathbb{R}^d$ with its Lebesgue measure, say) for $0<d<n$?
I'm sure the answer is $$\left(\displaystyle\sum_{1\leqslant n_1<\ldots<n_d\leqslant n}\prod_{k=1}^{d}a_{n_k}^2\right)^{1/2}.$$
As an easy special case $n=3$, $d=2$, the greatest area of an orthogonal projection of $[0,a]\times[0,b]\times[0,c]$ onto a plane is $\sqrt{a^2b^2+a^2c^2+b^2c^2}$.
An idea might be to consider the principle of "a projection of a convex hull of points is the convex hull of the projections of the points", and use Gram matrices. But I seem confused making a clean proof out of it.
This is by no means a complete answer, but Plucker Coordinates for a $k$-plane in $n$-space can be determined by taking a unit $k$-cube in that plane and projecting to each of the coordinate $k$-planes (there are $n \choose k$ of them) and determining the resulting $k$-dimensional volume in those planes. Those volumes are the plucker coordinates.
Thus the plucker coordinates for a line segment (which you can think of as a representation of a vector) in 3-space are exactly the $x$, $y$, and $z$-components of the vector. Similarly, the coordinates for a plane $P$ in 3-space come from taking a unit square in that plane and pushing it into each of the $yz$, $zx$, and $xy$ planes and measuring its area. As it turns out, the three resulting numbers $A,B,C$ are exactly the coefficients of $x, y, z$ in the plane-equation for $P$, which must therefore have the form $$ Ax + By + Cz = d $$ for some value $d$. (If we restrict to planes through the origin, then $d = 0$, of course).
Your question (if it were for a unit parallelipiped) would therefore be "what's the largest possible Plucker coordinate for my cube, in any conceivable orientation?"
I'll bet that this is the very sort of thing that Plucker examined, and maybe gave formulas for. So if I were trying to solve your problem, that's where I'd look.
(I was hoping to add the tag "Plucker Coordinates" to your question, but alas, there is no such tag.)