How large can the smallest generating set of a group $G$ of order $n$ be?

By a Theorem of Guralnick and Lucchini (which does require CFSG), if each Sylow subgroup of $G$ (ranging over all primes) can be generated by $r$ or fewer elements, then $G$ can be generated by $r+1$ or fewer elements. As noted in comments, if $G$ has a Sylow $p$-subgroup $P$ of order $p^{a}$, then $P$ can be generated by $a$ or fewer elements (and $a$ are needed if and only if $P$ is elementary Abelian). Hence if $|G|$ has prime factorization $p_{1}^{a_{1}}p_{2}^{a_{2}} \ldots p_{r}^{a_{r}}$ with the $p_{i}$ distinct primes, and the $a_{i}$ positive integers, then $G$ can be generated by $1 + {\rm max}(a_{i})$ or fewer elements.

(The result attributed to Guralnick and Lucchini was not a joint paper, rather a result proved independently at around the same time: references:

R. Guralnick, "A bound for the number of generators of a finite group, Arch. Math. 53 (1989), 521-523.

A Lucchini: "A bound on the number of generators of a finite group", Arch. Math 53, (1989), 313-317).

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.