Is any interesting question about a group G decidable from a presentation of G?

It seems to me that the analogue of Rice's theorem fails for finitely presented groups $G$ because of questions like: is the abelianization of $G$ of rank 3? The rank of the abelianization of any finitely presented $G$ can be computed by reducing the abelianization to normal form, so this (slightly) interesting question can be decided from the presentation of $G$.

Sorry I came to this question late. $F_q$ is undecidable from a presentation for each fixed $3\leq q<\infty$. First note that for finitely presented groups $F_q$ is equivalent to the homological finiteness condition $FP_q$ so it is enough to show this condition undecidable. This was done for $q =3$ in Section 5 of Cremanns, Robert; Otto, Friedrich, For groups the property of having finite derivation type is equivalent to the homological finiteness condition FP3. J. Symbolic Comput. 22 (1996), no. 2, 155–177 https://www.sciencedirect.com/science/article/pii/S0747717196900462. The same construction works for any fixed finite $q\geq 3$. They use essentially the same construction as the proof that Markov properties are undecidable but different details.

[This answers Reid's petition for an example in the comments, in answer form so as to be able to preview]

Stallings has given in [Stallings, John. A finitely presented group whose 3-dimensional integral homology is not finitely generated. Amer. J. Math. 85 1963 541--543. MR0158917] an example of a finitely presented group $G$ such that $H_3(G)$ is not finitely generated. It follows from this that the 3-skeleton of $BG$ is infinite. The group is $$G=\langle a,b,c,x,y:[x,a], [y,a],[x,b],[y,b],[a^{-1}x,c],[a^{-1}y,c],[b^{-1}a,c]\rangle.$$ Stallings' paper is characteristically short and beautiful, and the proof is a nice application of Mayer-Vietoris (!)