$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$I can improve Will Sawin's bound by a factor of $2$. Let $G$ be the group of $3 \times 3$ matrices with entries in $\ZZ/2^{\ell+1} \ZZ$ that are $\Id_3 \bmod 2$. This group has cardinality $N=2^{9\ell}$ and composition length $n = 9\ell$. I'll show that it has derived length is $\geq \lfloor \log_2 \ell \rfloor= \log_2 n + O(1) = \log_2 \log_2 N + O(1)$.
Let $U_{ij}(a)$ be the matrix $\Id_3 + 2^a e_{ij}$, where $e_{ij}$ is the matrix with a $1$ in position $(i,j)$ and a $0$ everywhere else. A quick computation checks that $$(U_{ij}(a), U_{jk}(b)) = U_{ik}(a+b)$$ for $i$, $j$, $k$ distinct. Thus, by induction, $U_{ij}(2^r)$ is in the $(r-1)$-st derived subgroup for $i \neq j$ and, in particular, the derived length is at least $\lfloor \log_2 \ell \rfloor$.