The general answer is $\log_2 n$. It is elementary to prove that this number suffices. Just choose $1 \ne g_1,g_2,g_3,\ldots \in G$ with $g_{i+1} \not\in G_{i} := \langle x_1,\ldots,x_i \rangle$, until $G_k=G$. Since each $G_i <G_{i+1}$ for $i<k$, we have $|G_{i+1}/G_i| \ge 2$, so $|G| = |G_k| \ge 2^k$.

But since an elementary abelian $2$-group requires that number of generators, this bound is best possible.

The general answer (as a function just of $n$, rather than of its factorization into primes) is $\log_2 n$. It is elementary to prove that this number suffices. Just choose $1 \ne g_1,g_2,g_3,\ldots \in G$ with $g_{i+1} \not\in G_{i} := \langle x_1,\ldots,x_i \rangle$, until $G_k=G$. Since each $G_i <G_{i+1}$ for $i<k$, we have $|G_{i+1}/G_i| \ge 2$, so $|G| = |G_k| \ge 2^k$.

But since an elementary abelian $2$-group requires that number of generators, this bound is best possible.