This is naive, but can you take the spherical warped product and apply the 4D case?

@MoisheKohan: I think you can arrange things so that the submanifold has fundamental group the expected edge group. Take an equivariant map to the tree induced by the HNN splitting over a finitely presented group, and make the preimage an embedded submanifold by transversality. If the preimage is not connected, tube components together to get a connnected preimage whose fundamental group surjects the edge group. Then do compressions, which keep the manifold connected because of the high dimension. This should terminate with a submanifold whose fundamental group is the edge group.

@HJRW does $ker\{ F \times F \to Z\}$ not split as an amalgamated product over a finitely presented subgroup? If that’s true, then I think your suggestion gives a counterexample. Maybe that follows from the classification of finitely presented subgroups of $F\times F$.

The stable or unstable foliation of an Anosov flow is a Reebless foliation. There are hyperbolic 3-manifolds without Reebless foliations, hence without Anosov flows and with exponential growth of the fundamental group. doi.org/10.1090/S0894-0347-03-00426-0

In fact it’s probably true in dim > 4 iff the fundamental group splits as an HNN extension over a finitely presented group.

If $-1<K<0$, then this follows from the generalized Margulis lemma. See Theorem 9.5 doi.org/10.1007/978-1-4684-9159-3 and the Cheeger-Gromov compactness theorem. mathoverflow.net/a/258518/1345 From compactness, one gets that the pinched nonpositively curved manifolds with volume bounded below are compact in the Hausdorff topology, and hence one gets a lower bound on injectivity radius everywhere. If the volume approaches zero, then the max injectivity radius approaches $0$. By 9.5, the fundamental group is virtually nilpotent, and hence not negatively curved, a contradiction.

I don’t think one can say anything like this without a bound on curvature. I think I can construct counterexample with the curvature going to $-\infty$.

@GeorgesElencwajg : the point of this answer was to give an overkill proof. To find a Riemann structure, you could put a Riemannian metric on it, giving it a conformal structure, then use that oriented conformal structures are equivalent to Riemann surface structure. Or use a pants decomposition to put a hyperbolic metric, hence Riemann surface structure.