I am curious about whether there is a nontrivial group homomorphism $\mathbb{Q} \to SL_n(\mathbb{Z})$ for some $n$. It's not hard to find such a homomorphism $\mathbb{Q} \to SL_2(\mathbb{Q})$; we can take the map given by $x \mapsto \left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right)$, but I don't see any obvious map into $SL_n(\mathbb{Z})$.
Another, weaker, question of interest is whether some $SL_n(\mathbb{Z})$ has an element $A \neq I$ for which $A$ has a $k$-th root in $SL_n(\mathbb{Z})$ for every $k$. In general, for a group $G$, the condition that $G$ has an element with $k$-th roots for every $k$ is strictly weaker than the condition that there is a nontrivial homomorphism $\mathbb{Q} \to G$, so these questions may have different answers.
No. The group $\Bbb{Q}$ and its nontrivial quotients are divisible and so non-residually finite while $SL_n(\Bbb Z)$ is residually finite.
The answer to the second question is also "no". Suppose a residually finite group $G$ has an element $g\ne 1$ such that for every integer $k>0$ there exists $g_k$ with $g_k^k=g$. Then there exists a finite group $H$ and a homomorphism $\phi: G\to H$ such that $\phi(g)\ne 1$. Let $|H|=k$. Then $\phi(g)=\phi(g_k)^k=1$, a contradiction.