If you want to apply Polya's theorem, you still have to enumerate, say, the number of all "cube parts" that are not identified by a permutation. Note also that I enumerate according to your comment on identifying by rotation and reflection not on your false description in the OP that thease are "topologoically isomorphic".
It is easier to count them directly than to use Polya theory.
Note also that I enumerate according to your comment on identifying by rotation and reflection not on your false description in the OP that these are "topologically isomorphic".
0 things: 1 possibility
1 vertex: 1 possibility
2 vertices: 3 possibilities
3 vertices in a plane: 1 possibility
3 vertices not in a plane (which implies that two of them are diagonally opposite or they are the three neighbours of one vertex): 2+1=3 possibilities
4 vertices in a plane: 1 possibility
4 vertices not in a plane:
5 vertices (=3 vertices): 4 possibilities
6 vertices (=2 vertices): 3 possibilities
7 vertices (=1 vertex): 1 possibility
8 vertices: 1 possibility
1 edge and some vertices (all choices of vertices + symmetric choices of vertices divided by 2): (2^6+2^4)/2=40 possibilities
2 edges in a plane and some vertices (number of vertices 0,1,2,3,4): 1+1+3+1+1=7
2 parallel edges not in a plane and some vertices: 1+1+3+1+1=7
2 non-parallel edges and some vertices: 1+2+4+2+1=10
3 parallel edges and some vertices: 1+1+1=3 possibilities
3 edges (not all parallel) and some vertices: 1+1+1=3 possibilities
4 edges (all parallel or not): 2 possibilities
1 face and some vertices: 1+1+2+1+1=6 possibilites
1 face, 1 edge and some vertices: 1+1+1=3 possibilities
1 face and 2 edges: 1 possibility
2 faces: 1 possibility
Now add everything.