It is well-known that the ends of a finite-volume hyperbolic manifold are warped products $$(0,\infty)\times\_f T$$
for some euclidean manifold $T$ and $f(t)=e^{-t}$.


Question: Is there a similar result for rank 1 symmetric spaces (or even for more general negatively curved manifolds)?



Cusps as warped products
It may be not the answer you are hoping for but still: one can show (short tensor calculations) that, even locally, the metric of an irreducible 
 symmetric space different from the space of constant curvature 
can not be a warped product metric. 


Now, concerning arbitrary negatively curved metrics: I do not really understand what you would like to know; let me give three answers on different versions of your question and if possibly please follow the rules and write the motivation for your question 


(1) of course a generic metric, even negatively curved, is not a warped product metrics. 


(2) Of course there exist examples of negatively curved metrics such that they are warped products near cusps -- just take a manifold of constant negative curvature and slightly perturb it locally, ``far from cusp''. 


(3) Warped structure relates to the existence of a nontrivial solution for the equation 
$$
\textrm{Trace-free-part-of}(\nabla \nabla u)=0 \ \ \ (\ast)$$
which is a system of linear overdetermined PDEs on the unknown function $u$. In a neighborhood of almost every point, metrics admitting a solution of $(\ast)$ are locally warped product, and warped product metrics locally admit a solution of such equation. 


On can show that complete negavely curved manifolds of finite volume do not have nontrivial solutions of $(\ast)$. 


