Se Does $SL_3(R)$ embed in $SL_2(R)$? for a related discussion.

That any homomorphism $\varphi\colon SL_n(\mathbb{Z}) \to GL_{n-1}(\mathbb{Z})$ is trivial can be seen as follows.

By Margulis' normal subgroup theorem, either the kernel of $\varphi$ is finite (in which it is trivial or the center of $SL_n(\mathbb Z)$, which is $\pm I_n$) or the image of $\varphi$ is finite.

In the second case, the kernel of $\varphi$ contains a congruence subgroup, so $\varphi$ factors through a group $SL_n(\mathbb Z/m)$ for some $m \in \mathbb Z$. But then $\varphi$ gives rise to a nontrivial representation of this group of degree $n-1$, which is impossible.

In the first case, note first that the subgroup $U_n(\mathbb{Z}) \leq SL_n(\mathbb{Z})$ of upper triangular matrices is nilpotent of class $n-1$, and it is a nice exercise to see that any subgroup $U'\leq U$ of finite index is also nilpotent of class $n-1$.

Now if $\varphi$ were nontrivial, it would inject $U$ into $GL_{n-1}(\mathbb{C})$ since $U$ intersects the center of $SL_n(\mathbb{Z})$ trivially. Then $U$ contains a finite index subgroup $U'$ such that the Zariski closure of $\varphi(U')$ is connected: take $U'$ to be the preimage under $\varphi$ of the connected component of the identity of $\overline{\varphi(U)}$.

By the Lie-Kolchin theorem, it follows that $\varphi(U')$ is conjugate to a subgroup of the upper triangular matrices, which is a contradiction since any such subgroup has nilpotency class at most $n-2$.