Number of stereoisomers of polysubstituted cubane The molecule given below is a substituted cubane.

a) How many stereoisomers exist for this molecule?
b) How many pairs of enantiomers are possible?

How do I calculate the number of stereoisomers? Since each carbon is linked to four different groups, can I claim that all are chiral and, therefore, $2^8$ stereoisomers are possible?
If that is true, since there is no element of symmetry, can I say that $2^4$ enantiomeric pairs exist?
Edit: Looking at the comments clears things up but brings forth another question. If, say, iodine was present at any other position that would not lead to a tetrahedron, what would happen?
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.