The answer is no, even in higher rank groups. For example, take $\Gamma = SL_3({\mathbb Z})$. If such a $\Sigma $ existed for any $n$, then its completion in the profinite (same as congruence) completion of $\Gamma$ would have this property that every element in $\Sigma$ would be an $n$-th power. But the completion of $\Sigma$, by strong approximation, is open. So for almost all primes, $SL_3({\mathbb Z}_p)$ would have this property. But you can easily cook up a prime $p$ and elements in $SL_3({\mathbb F}_p)$ which are not $n$-th powers.

To see this, get large primes such that $p^3 \equiv 1\quad (mod\quad n)$ but $p-1$ is not divisible by $n$. Choose a cubic extension of ${\mathbb F}_p$; the group of norm one elements in the cubic extension is cyclic of order $\frac{p^3-1}{p-1}$. Hence the $n$-th power map on this subgroup of $SL_3({\mathbb F}_p)$ is not surjective; but any $n$-th root of a generator for this subgroup already lies in this subgroup.

[Edit] To summarise: for any lattice in a semi-simple group, and for any integer $n\geq 2$, the image of the $n$-th power map never contains a finite index subgroup. The argument using strong approximation works if the lattice has number field entries (this holds for all lattices except those in $SL(2,{\mathbb R})$) . I think that it is true for any Zariski dense finitely generated subgroup of a semi-simple group, using strong approximation due to Nori (and Weisfeiler).

semisimple(real or $p$-adic) Lie groups. – Yves Cornulier Feb 24 '13 at 21:46