Is it true that $\lim_{n \to \infty} {(P(\forall i,j\leq n \text{ } [X_i, X_j] = e))}^{\frac{1}{n}} = P(X_1 \in Z(G))$?

I believe $S_3$ is a counterexample. Let $H< S_3$ be the unique subgroup of index $2$. Then $H$ is abelian, and for every $n$ we have $$P(\forall i,j:[X_i,X_j]=e)^{1/n} \geq P(\forall i: X_i\in H)^{1/n}=\frac{1}{2}.$$ This means the limit on the left (if it exists) is at least $1/2$. On the other hand, $Z(S_3)=\{e\}$, so $P(X_1\in Z(S_3))=P(X_1=e)=1/6$.

For general $G$, a similar argument should show that $$ \liminf_{n\to\infty}P(\forall i,j\leq n:[X_i,X_j]=e) \geq \max_{\substack{H\leq G\\H\text{ abelian}}}\frac{1}{[G:H]}. $$

For finite $G$, the bound given in Julian Rosen's answer is the exact limit, i.e. I claim that $$ \mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n}\to \max_{\text{abelian }H \le G}\frac{1}{[G:H]} = \max_{\text{abelian }H \le G}\mathrm P(X_1 \in H). \tag{$\ast$} $$

Indeed, denoting by $\mathcal A$ the collection of abelian subgroups of $G$, $$ \limsup_{n\to \infty} \mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n} \le \limsup_{n\to \infty}\biggl(\sum_{H\in \mathcal A} \mathrm P(\forall i\le n,\ X_i\in H)\biggr)^{1/n} =\\ = \limsup_{n\to \infty}\biggl(\sum_{H\in \mathcal A} \mathrm P(X_1\in H)^n\biggr)^{1/n} = \max_{H\in \mathcal A} \mathrm P(X_1\in H). $$

Since by Julian Rosen's answer, $$ \liminf_{n\to \infty} \mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n}\ge \max_{H\in \mathcal A}\mathrm P(X_1 \in H), $$ we arrive at $(\ast)$.

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In order for (a modification of) the argument to work, it is enough to assume that for some $n\ge 1$, $\sum_{H\in \mathcal A^*} \mathrm P(X_1\in H)^n<\infty$, where $\mathcal A^*$ is the collection of all maximal abelian subgroups of $G$.