Auslander (sorry, restricted access link) proved that the automorphism group of any polycyclic group is residually finite, linear and finitely presented. As mentioned by Steve in the comments, it extends to virtually polycyclic groups (se the book Polycyclic Groups by Segal).

For $\mathrm{Out}(G)$, we have to mod out by the group of inner automorphisms, which is f.g. virtually abelian. This preserves being finitely presented.

It also follows that $\mathrm{Out}(G)=\mathrm{Aut}(G)/\mathrm{N}$ is residually finite: indeed, let $N$ be a characteristic abelian subgroup of finite index in $G$. The profinite closure $A$ of $\mathrm{Inn}(N)$ is finitely generated abelian: indeed, considering the action of $G$ on $N$, we see that $G$ is an extension of two groups both linear over $\mathbf{Z}$, so has all its abelian subgroups finitely generated. By construction, $G/A$ is residually finite. Since $A/\mathrm{Inn}(N)$ is f.g. and residually finite, it follows that $G/\mathrm{Inn}(N)$ is residually finite. By modding out by a finite subgroup we then get $\mathrm{Out}(G)$, which is residually finite.

I'm not quite sure about linearity, but I think that we could need some slightly more refined result about $\mathrm{Aut}(G)$, saying that it has a linear representation over $\mathbf{Z}$ so that the image of $\mathrm{Inn}(G)$ is virtually unipotent, to pass linearity to the quotient. Or there's a trick to avoid this (and maybe an available reference as well). [One more edit: Steve pointed out that Out of virtually polycyclic groups are indeed linear over $\mathbf{Z}$: Wehrfritz, Two Remarks on Polycyclic Groups, Bull. LMS 26, 543--548]

[Edit: my previous arguments were too flawed to be maintained.]