I answer here the case of $\mathrm{SL}_2$, with actually an argument carrying over $\mathrm{SL}_n$. Namely, $\mathrm{SL}_n(\mathbf{Z})$ is maximal among lattices in $\mathrm{SL}_n(\mathbf{R})$. Indeed, let $\Lambda$ be an overgroup of finite index.

First assume that $\Lambda\subset\mathrm{SL}_n(\mathbf{Q})$. Then there is a bound on denominators in $\Lambda\subset\mathrm{M}_n(\mathbf{Q})$. It follows that the subgroup of $\mathbf{Q}^n$ generated by the $g\mathbf{Z}^n$ when $g$ ranges over $\Lambda$, is a lattice. So $\Lambda$ preserves this lattice and hence is conjugate to a subgroup of $\mathrm{SL}_n(\mathbf{Z})$. But since the automorphism group of $\mathrm{SL}$ preserves the volume, this implies that $\Lambda=\mathrm{SL}_n(\mathbf{Z})$.

Now we have to show that $\Gamma$ is contained in $\mathrm{SL}_n(\mathbf{Q})$. To show this, let us first understand orbits in $\mathbf{R}^n$ of finite index subgroups $\Xi$ of $\mathrm{SL}_n(\mathbf{Z})$. Start from a nonzero vector $v$, we wish to describe the closure $X$ of the additive subgroup generated by the orbit $\Lambda v$. Note that $X$ is the closure of $Av$, where $A$ is the $\mathbf{Z}$-subalgebra generated by $\Gamma$. For some $m\ge 1$, all elementary matrices $e_{ij}(m)=I+E_{ij}(m)$ belong to $\Xi$, and hence $E_{ij}(m)$ belongs to $A$. So, denoting by $(e_i)$ the canonical basis, $mv_je_i$ belongs to $X$ for all $i\neq j$. If the $v_j$ and $v\in X$ for $j\neq i$ generate a dense subgroup of $\mathbf{R}$, we deduce that the line $\mathbf{R}e_i$ is contained in $X$, and reiterating, we see that $\mathbf{R}e_j$ is contained in $X$ for all $j$ and hence that $X=\mathbf{R}$. We can do this as soon as $n\ge 3$ and $v$ is not scalar multiple of a rational vector. If $n=2$, assuming that $v$ is not multiple of a scalar vector, we can assume $v=(1,t)$ with $t$ irrational. Then $X$ contains the lattice $t\mathbf{Z}\times\mathbf{Z}$. Clearly $v$ has infinite order modulo this lattice, so the closure of $\mathbf{Z}v$ modulo this lattice is non-discrete, hence $X$ is not discrete, thus contains a line. Applying either $e_{12}(m)$ or $e_{21}(m)$ to this line yields another line and hence $X=\mathbf{R}^2$.

To summarize, we have shown that for every $v$ that is not scalar multiple of a rational vector, the subgroup generated by $\Xi v$ is dense. Thus, the lattices preserved by $\Xi$ are all scalar multiples of rational lattices (i.e., contained in $\mathbf{Q}^n$).

If $g\in\Lambda$, there exist two finite index subgroups $\Xi_1$ and $\Xi_2$ of $\mathrm{SL}_n(\mathbf{Z})$ such that $g\Xi_1=\Xi_2$. From the previous fact, we deduce that $g$ maps rational lattices to scalar multiples of rational lattices. Thus we can write $g=\lambda g'$ with $g'$ a rational matrix. The determinant condition implies that $1=\lambda^n\det(g)$, and thus $\lambda^n$ is rational. Also, $\lambda$ is unique in $\mathbf{R}^*/\mathbf{Q}^*$ and $g\mapsto \lambda=\lambda_g$ is a homomorphism into this group. The kernel is the intersection $\Lambda\cap\mathrm{SL}_n(\mathbf{Q})$, which has been shown to be $\mathrm{SL}_n(\mathbf{Z})$. So $\mathrm{SL}_n(\mathbf{Z})$ is normal in $\Lambda$.

So now we need to understand the normalizer of $\mathrm{SL}_n(\mathbf{Z})$. Using that $\mathrm{SL}_n(\mathbf{Z})$ is transitive on primitive elements in $\mathbf{Z}^n$, we see that the lattices preserved by $\mathrm{SL}_n(\mathbf{Z})$ are precisely the scalar multiples of $\mathrm{Z}^n$. So if an element normalizes it and preserves the volume, it has to preserve $\mathrm{Z}^n$. This shows that $\mathrm{SL}_n(\mathbf{Z})$ is self-normalized in $\mathrm{SL}_n(\mathbf{R})$, and finally $\Lambda=\mathrm{SL}_n(\mathbf{Z})$.

A similar argument shows that every lattice containing a finite index subgroup of $\mathrm{SL}_n(\mathbf{Z})$ is actually contained in a conjugate of $\mathrm{SL}_n(\mathbf{Z})$ by some rational matrix.