One can show that a cube is not even negligible using (hyper)rectangles in your definition of "negligible".

Indeed, given a cover with rectangles (w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a rectangle is $[a,b]$ there are no rectangles whose side (along the same direction) has an endpoint in $(a,b)$. Now two rectangles are either disjoint or identical (up to borders) and, discarding duplicate rectangles, the sum of the volumes is the volume of the cube.