Cubes trivially tile the $3D$ space, but I wonder if this is also true for their duals: octahedrons?
For $2D$ case, the answer is yes: The dual of square can tile the floor, but for $d \geq 3$ it is not super obvious. Notice that for higher dimensions, you can consider the dual to be the unit ball with $L1$-norm.
What we can do is stack an appropriate mixture of regular octahedra and regular tetrahedra. Start with a regular octahedron. Add two regular tetrahedra whose edges are congruent to those of the octahedron, matching one face of each tetrahedron to one of two opposite faces of the octahedron. Now you have a solid whose dihedral angles come in supplementary pairs thus enabling a periodic stacking.