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

Indeed, given a cover with rectangles (w.l.o.g. open up to enlarging them a bit, w.l.o.g. finite by compactness of the cube, and w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a rectangle is $[a,b]$ (or $(a,b)$, $[a,b)$, $(a,b]$) there are no rectangles whose side (along the same direction) has an endpoint in $(a,b)$.

Now replace the rectangles with their closure. 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.

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

Indeed, given a cover with rectangles (w.l.o.g. initially open up to enlarging them a bit, w.l.o.g. finite by compactness of the cube, and w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a rectangle is $[a,b]$ (or $(a,b)$, $[a,b)$, $(a,b]$) there are no rectangles whose side (along the same direction) has an endpoint strictly between $a$ and $b$.

Now replace the rectangles with their closure. 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.

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.

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

Indeed, given a cover with rectangles (w.l.o.g. open up to enlarging them a bit, w.l.o.g. finite by compactness of the cube, and w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a rectangle is $[a,b]$ (or $(a,b)$, $[a,b)$, $(a,b]$) there are no rectangles whose side (along the same direction) has an endpoint in $(a,b)$.

Now replace the rectangles with their closure. 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.