This is a reference request, since the answer is probably well known, but I could not find it.

Given a finitely generated group $\Gamma$ with a generating set $S$, define the word norm $l = l_S : \Gamma \rightarrow \mathbb{N}$ to be $$l (g) = \min \lbrace k : \exists s_1,...,s_k \in S, g = s_1 ... s_k \rbrace ,$$ i.e., $l(g)$ is the distance between $e$ and $g$ in the Cayley graph of $\Gamma$ (w.r.t the generating set $S$).

My question: Let $\Gamma = \rm SL_3 (\mathbb{Z})$ (with some finite generating set). For $1 \leq i,j, \leq 3, i \neq j$ and $m \in \mathbb{Z}$, denote $e_{i,j} (m)$ to be the elementary matrix with $1$'s along the main diagonal, $m$ in the $(i,j)$-entry and $0$ in all other entries. What can one say about the growth rate of $l (e_{i,j} (m))$?

My naive attempt for an answer gives me $l (e_{i,j} (m)) = O (\log^3 (m))$:

For convenience, we fix the generating set $S = \lbrace e_{i,j} (\pm 1), e_{i,j} (\pm 2) : 1 \leq i,j, \leq 3, i \neq j \rbrace$.

Using commutator (Steinberg) relations $$ e_{i,j} (2^{2^{r+1}}) = [e_{i,k} (2^{2^{r}}), e_{k,j} ( 2^{2^{r}})]$$ it is not hard to show by induction on $r$ that $l (e_{i,j} (2^{2^r})) \leq 4^{r} $.

Again by the commutator relations, it follows that for every $2^r \leq d < 2^{r+1}$ it holds that $l (e_{i,j} (2^{d})) \leq 4^{r+1} \leq 4 d^2$.

Thus it follows that for every $2^d \leq m < 2^{d+1}$,

$$l (e_{i,j} (m)) \leq \sum_{t =0}^d l (e_{i,j} (2^t)) \leq 4 \sum_{t =0}^d t^2 = 4 \frac{d (d+1) (2d +1)}{6} = O (\log^3 (m)).$$

As noted above - this computation is quite naive. Are there better known results?