This addresses the second question "What is known about finite subgroups of $SU(n)$". 
A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$
such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4
of [Thurston's book][1]). 
Thus, there is a normal abelian subgroup of index at most $m!$. So one may make
a statement: there are finitely many finite groups so that any finite subgroup of 
$SU(n)$ is an abelian extension (of rank at most $n-1$) of one of these finitely many groups. It would
be quite interesting to obtain an estimate of the function $m(n)$,
which should be possible by giving an effective proof of Margulis' theorem. 
I did a literature search once to see if anyone had attempted this, but I didn't
find anything, and I would be curious if someone knows something. 

Addendum: Working backwards from Weisfeiler's paper referenced in Keivan's comment, I found [a result of Collins][2] implies that a finite linear subgroup of $GL(n,C)$ has an abelian normal subgroup of  index at most $(n+1)!$ when $n\geq 71$ (and gives the bound for all $n$). Since finite subgroups of $GL(n,C)$ are conjugate into $U(n)$, this bound works for $SU(n)$. See also [Collins paper][3] on primitive representations, which has some historical discussion of this problem.  


  [1]: http://math.berkeley.edu/~ianagol/276.S10/bookdraft.pdf
  [2]: http://www.ams.org/mathscinet-getitem?mr=2334748
  [3]: http://www.ams.org/mathscinet-getitem?mr=2381807