Is $\mathbb{H}^n$ quasi-isometric to a leaf of a codimension 1 foliation of a compact manifold? If we extend the action of $\pi_1(\Sigma_g), g\geq 2,$ from $\mathbb{H}^2$ to its boundary $\partial_{\infty}\mathbb{H}^2=S^1$, the surface bundle corresponding to this action of $\pi_1(\Sigma_g)$ on $S^1$ has a leaf that is isometric to $\mathbb{H}^2$.  Hence $\mathbb{H}^2$ is even isometric to a leaf in a codimension 1 foliation of a compact manifold.
For $n>2$ the above argument fails, and hence my question in the title:  Is it known whether for every $n$ $\mathbb{H}^n$ is quasi-isometric (or even isometric) to a leaf of a foliation of a compact (n+1)-manifold?
 A: One may generalize your 3-dimensional example to all dimensions in some sense. 
There are compact hyperbolic $n$-manifolds which have orderable fundamental group, since the fundamental groups embed in a right-angled Artin group (which is in fact bi-orderable), by a result of Haglund and Wise. Let $M=\mathbb{H}^n/\Gamma$ be such a manifold, with faithful action $\Gamma \to Homeo^+(\mathbb{R})$ (this is equivalent to being orderable), and consider the twisted product $(\mathbb{H}^n \times \mathbb{R})/ \Gamma$, where $\Gamma$ acts diagonally. The foliation $\mathbb{H}^n\times \{x\}$ naturally descends to a foliation on the quotient. This manifold is homeomorphic to $M\times \mathbb{R}$, so embeds in $M\times [-\infty,\infty]$, where the boundary $M\times \{\pm\infty\}$ are leaves of the foliation. There is a point $x\in\mathbb{R}$ on which $\Gamma$ acts faithfully (in particular, the orbit induces the total ordering on $\Gamma$) if the action is well-chosen (see the proof of Theorem 6.8 in Ghys' paper). Thus,  $(\mathbb{H}^n \times \{\Gamma x\})/\Gamma \cong \mathbb{H}^n$, so there is a leaf of the foliation which is a copy of $\mathbb{H}^n$. Since it is uniformly close to $\tilde{M}\times \{\infty\}$ in the universal cover, it will be quasisometric to $\tilde{M}=\mathbb{H}^n$. Glue $M\times{-\infty}$ to $M\times \{\infty\}$ to  obtain a closed manifold $M\times S^1$ with the desired property (really, we are embedding $Homeo^+(\mathbb{R})\subset Homeo^+(S^1)$ by the 1-point compactification, and taking the corresponding twisted product). 
A: Let me add a comment. One knows that a codimension-one foliation without leafwise holonomy must be given by a Z^n action, by a result of Sacksteder
Sacksteder, Richard
Foliations and pseudogroups.
Amer. J. Math. 87 1965 79–102. 
So, if the foliation is defined by a group action on the circle, and the orbits are QI to a hyperbolic space, then there must be leaves with holonomy. This means there is some point on the circle for which there is a non-trivial element of the group that fixes the point. Maybe this can be used to prove that no such actions exist. In any case, it proves that not all leaves are QI to H^n.
