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$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $$\geq \lfloor \log_2 \ell \rfloor= \log_2 n + O(1) = \log_2 \log_2 N + O(1).$$

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 $\lfloor \log_2 \ell \rfloor$.

Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (Warning: I tried to check this, and I got $n = 4^r-1 = 3 \log_{24} N$. That doesn't affect the asymptotics, but it gives me cause for caution. The $d=3r$ looks plausible to me, but I didn't fully check it.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $$\geq \lfloor \log_2 \ell \rfloor= \log_2 n + O(1) = \log_2 \log_2 N + O(1).$$

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 $\lfloor \log_2 \ell \rfloor$.

Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (I have partially checked these numbers; I confirm $n = 4 \log_{24} N$ and I find $d=3r$ very plausible, but haven't fully checked it. My earlier comment claiming that Glasby's $n$ looked wrong was mistaken.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $\lfloor \log_2 \ell \rfloor$.

Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (I haven't checked this.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $$\geq \lfloor \log_2 \ell \rfloor= \log_2 n + O(1) = \log_2 \log_2 N + O(1).$$

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 $\lfloor \log_2 \ell \rfloor$.

Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (Warning: I tried to check this, and I got $n = 4^r-1 = 3 \log_{24} N$. That doesn't affect the asymptotics, but it gives me cause for caution. The $d=3r$ looks plausible to me, but I didn't fully check it.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $\lfloor \log_2 \ell \rfloor$.

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$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 $\lfloor \log_2 \ell \rfloor$.

Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (I haven't checked this.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$