Number of stereoisomers of polysubstituted cubane
I hope I am not too late.

All carbons are $\mathrm{sp^3}$-hybridised and three bonds of every carbon are fixed (one corner shares three bonds), so fourth one has only one place in space to maintain an $\mathrm{sp^3}$ tetrahedron. This one is optically active, so this and its enantiomer are the only two possible isomers.

Now you get why we don't use the $2^n$ formula. And similar thing happens in a bridgehead position.