This version works for all parallelepipeds, not only rectangular ones:
If you replace each parallelepiped by all points that have distance at most $\varepsilon$ to a point in the parallelepiped, you can still place the smaller inside the bigger one. In particular, the smaller object has a smaller volume. The volumes are composed of
We divide up the extended parallepipeds by extending the planes corresponding to the six faces. This gives the volume of the original box, solid in the volume center, parallelepipeds of some boxes glued to height $\varepsilon$ on top of each face, the volume of some partial cylinders glued to each edge (with slanted parallel ends) of radius $\varepsilon$ and length the volume of some corresponding edge, and partial spheres glued to of radius $\varepsilon$ around each vertex.
The partial spheres add up to exactly one whole sphere simply by translation. The partial cylinders corresponding to parallel edges add up to one whole cylinder by translation.
Now let $\varepsilon$ tend to infinity (yes, really). The term with $\varepsilon^3$ has some fixed constant that does not depend on the edges (calculated comes from the portion of the sphere of radius $\varepsilon$ and does not depend on the number of vertices)parallelepiped at all, . The term with coefficient of $\varepsilon^2$ comes from the cylinders and is some fixed constant times clearly the sum of the edges times some constant.
So, for the inequality to be valid the sum of the edges of the smaller box must be smaller than the sum of the edges of the larger box.
I don't see any problem with the generalization to higher dimension.