I accepted Todd's answer because it was very helpful, but I'm going to write this answer because I figured out a *very* useful heuristic way to think about this (although I think that actually proving the equivalence wouldn't be very tough (or worthwhile)).

There is a very nice geometric description that makes the combinatorics of the problem much much clearer: Imagine that all of the maps involved are just the inclusion $0\to \mathbf{R}$. Then the box product of $n$ of these maps is the inclusion $X\hookrightarrow \mathbf{R}^n$, where $X$ is the union of all codimension 1 hyperplanes spanned by the axes. We can associate a commutative diagram with this description, which is generated by the different ways to glue over different maps, and this exactly represents the general problem. The reason why embedding in $\mathbf{R}^n$ in this way is so effective is that any two of the relevant objects are glued along their intersection. The embedding encodes the explicit combinatorial information so you don't need to worry about it.

In particular, without this picture (or a similar one) in mind, it will very quickly get extremely difficult to deal with box products of anything more than a few maps, since there are already tons of ways to glue together just three planes passing through the origin if you do it piece-by-piece. This is because using "pushout notation" as above, we not only have to keep track of the pieces we're gluing together, but also what we're gluing them along. If you don't understand the geometric picture, you're essentially proving it by brute force by drawing every single possible pushout path (see Todd's answer; indeed if there were more than three maps, the commutative diagram would be four dimensional) to the required map (this is why I was unable to draw the diagram, since it is at least this complicated, but possibly moreso):

I hope people learn from my mistakes!