A very wide array of properties are compatible with these hypotheses.  Burger--Mozes famously gave examples of infinite simple groups of this form.  Earlier, Wise and Bhattacharjee had independently given examples of such groups with no proper finite quotients.  In particular, these groups have no non-trivial nilpotent quotients.

Let's call such an example $S$.  Bridson and I used $S$ to prove the following result (see Proposition 7.5 of [arXiv:1401.2273][1]).

>**Proposition 1:** Given any finitely presented group $G$ there exists a compact, non-positively curved square complex $X$ with a surjection
>
>$\eta:\pi_1X\to G$
>
>such that $\eta$ induces an isomorphism on profinite completions.

*Sketch proof:* Take a finite presentation complex $Y$ for $G$.  Replace each 2-cell of $Y$ with an annulus, and glue the other end of the annuli to npc square complexes with fundamental groups $S$.  If this is done carefully, the resulting complex $X$ is a non-positively curved square complex.  The map $\pi_1X\to G$ is obtained by killing the copies of $S$ inside $\pi_1X$. *QED*

Notice that, since $S$ is also has no nilpotent quotients, one obtains in the same way:

>**Proposition 2:** Given any finitely presented group $G$ there exists a compact, non-positively curved square complex $X$ with a surjection
>
>$\eta:\pi_1X\to G$
>
>such that $\eta$ induces an isomorphism on pro-nilpotent completions.

[That is, every nilpotent quotient of $\pi_1X$ factors through $\eta$.]

The group $\pi_1X$ acting on the universal cover $\widetilde{X}$ satisfies the hypotheses of the question, and the conclusion is that the nilpotent quotients of such groups are exactly as general as the nilpotent quotients of an arbitrary finitely presented group.

So nothing can be said about the nilpotent quotients of such a group without some additional information.


  [1]: http://arxiv.org/abs/1401.2273