$SL_n(\mathbb Z)$ is generated by elementary matrices. In general let $E_n(R)$ be the subgroup of $SL_n(R)$ generated by elementary matrices. Let's show that if $m\lt n$ and $n\ge 3$ then the only homomorphism $E_n(R)\to GL_m(F)$ is the trivial one, if $F$ is a field. Wlog $F$ is algebraically closed.

Let $T_n(R)\subset E_n(R)$ be the group of strictly upper-triangular matrices. It is generated by elementary matrices $e_{ij}(a)$ where $1\le i\lt j\le n$ and $a\in R$. Let's first show that every homomorphism $h:T_n(R)\to GL_m(R)$ takes $e_{1n}(a)$ to a scalar matrix. This is by induction on $n$. The element $e_{1n}(a)$ is central in $T_n(R)$, so the image of $h$ is contained in the centralizer of $h(e_{1n}(a))$. If $h(e_{1n}(a))$ is not a scalar, then by conjugating we can assume the image of $h$ is contained in $GL_{m_1}(F)\times GL_{m_2}(F)$ for some positive $m_1$ and $m_2$ adding up to $m$. (EDIT: I have changed the next few sentences to correct an error.) Or rather, that the image of $h$ is contained in the subgroup of $GL_m(F)$ preserving $F^{m_1}\times 0\subset F^{m_1}\times F^{m_2}=F^m$. By induction on $n$ the images of $e_{1,n-1}(a)$ under the resulting homomorphisms to $GL_{m_1}(F)$ and $GL_{m_2}(F)$ are central. (To begin the induction, if $n=3$ then $m_1$ and $m_2$ are both $1$ so this step still is OK.) So $h$ takes $T_n(R)$ into some conjugate of $T_m(F)$. Since $n>m$, it follows that $e_{1n}(a)$, being an $(n-1)$-fold commutator in $T_n(R)$, to the identity after all.

In particular a homomorphism $E_n(R)\to GL_m(F)$ must take $e_{1n}(a)$ to a scalar matrix. Of course more generally it must take $e_{ij}(a)$ to a scalar matrix. Since $n\ge 3$, we can express $e_{ij}(a)$ as the commutator of $e_{ik}(a)$ and $e_{kj}(1)$, and it follows that $e_{ij}(a)$ goes to the identity.