Here's a proof of finiteness.

First, it's a general fact that for any centerless directly indecomposable groups $A_1,\dots,A_k$, any automorphism permutes the $A_i$. References are welcome; one (in French), probably much too recent, is Proposition 2 p9 here (2006 paper by myself and Pierre de la Harpe). In particular, if all $A_i$ have finite Out, so does the product.

This applies to $\Lambda=\prod_{i=1}^m\mathrm{PSL}_{n_i}(\mathbf{Z})=[\Gamma,\Gamma]/Z(\Gamma)$.

It follows that some finite index subgroup of $\mathrm{Aut}(\Gamma)$ acts by inner automorphisms on $\Lambda$. To conclude, one has to show that the subgroup $F$ of $\mathrm{Aut}(\Gamma)$ acting as the identity on $\Lambda$ is finite. To show this, it is enough to show that the following finite index subgroup of $F$ is itself finite: the kernel $F'$ of the $F$-action on $Z(\Gamma)\times (\Gamma/[\Gamma,\Gamma])$. The $F'$-action on $[\Gamma,\Gamma]$ consists is by automorphisms of the form $g\mapsto gs(g)$ where $s$ are homomorphisms $[\Gamma,\Gamma]\to Z(\Gamma)$. Since $Z(\Gamma)$ is finite and $[\Gamma,\Gamma]$ is finitely generated, the set of such homomorphisms is finite. Hence some finite index subgroup $F''$ of $F'$ acts trivially on $[\Gamma,\Gamma]$. In turn, the $F''$-action on $\Gamma$ is by automorphisms that are the identity on the finite index subgroup $[\Gamma,\Gamma]$, and have the form $x\mapsto xu(x)$ where $u$ is a map from $\Gamma$ to $[\Gamma,\Gamma]$. Since it's an automorphism, we have $u(xy)=y^{-1}u(x)yu(y)$ for all $x,y$, and $u=1$ on $[\Gamma,\Gamma]$. Combining, we see that $u$ factors through $\Gamma/[\Gamma,\Gamma]$. So $u(xy)=u(yx)$ for all $x,y$. Taking $y$ in $[\Gamma,\Gamma]$ then yields $u(x)=y^{-1}u(x)y$ for all $x$; thus $u$ takes values in $Z([\Gamma,\Gamma]$, which in our case is central in $\Gamma$. So $u$ is a homomorphism $\Gamma\to Z(\Gamma)$. Again the set of such homomorphisms is finite, and we have proved finiteness of $F''$, hence of $F$.