It is not too hard to prove that if $G$ is a non-abelian finite simple group and $A \leqslant G$ is an abelian subgroup, then $|A| \leq \sqrt{|G|}$ (this can be improved to $\ll |G|^{1/3}$ using the classification). There exists a similar bound for $G$ being perfect?
In particular, can the index $[G:A]$ be fixed while $G$ grows in this case?
There is no such bound, and yes $|G:A|$ can be fixed while $|G|$ grows.
Let $S$ be a finite simple group, and $M$ a faithful irreducible module for $S$ over a finite field. For example, we could take $S = A_5$ and $M$ the $4$-dimensional deleted permutation module over the field of prime order $p$ for any $p \ne 5$ (or the $3$-dimensional module when $p=5$).
Now let $G$ be the semidirect product $A \rtimes S$, where $A$ is the direct sum of $k$ copies of $M$ for any $k > 0$. Then $G$ is perfect with abelian normal subgroup $A$ of order $|M|^k$ with $|G:A| = |S|$.