Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components? Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components?  The only constructions I could find have one boundary component.  A reference would be appreciated.
 A: A beautiful construction (relying on 2-dimensional hyperbolic circle packings) of hyperbolic 3-manifolds with many totally geodesic boundary components can be found in the paper
Totally geodesic boundaries are dense in the moduli space.
J. Math. Soc. Japan 49 (1997), no. 3, 589--601.
by Fujii and Soma.
A: If you don't care about explicitly constructing the metric, you can do this:
Let $N$ be a closed orientable $3$-manifold.  Let $\Gamma$ be any finite graph such that each component has negative Euler characteristic.  A theorem of R. Myers (in his "Excellent 1-manifolds" paper) gives you an embedding of $\Gamma$ into $N$ such that $M=N \setminus \mathrm{nbhd}(\Gamma)$ is irreducible acylindrical atoroidal and boundary incompressible.
It then follows from Thurston's geometrization theorem for Haken manifolds that $M$ admits a hyperbolic metric with geodesic boundary.
A: A combinatorial way to construct examples consists of taking any (topological) ideal triangulation with k ideal vertices where all edges have valence >=7 (i.e. each edge meets at least 7 tetrahedra, counted with multiplicity: if you take at least 6 you might find toric cusps). After removing an open star of the vertices you get a 3-manifold with k boundary components which admits such a metric. 
(To prove this, see http://arxiv.org/abs/math/0402339, or prove that there cannot be any normal surfaces with non-negative Euler characteristic in such a triangulation and use geometrization.)
You can constrcut concretely the metric if all edges have the same valence v>=7: in this case you realize every tetrahedron as a regular truncated hyperbolic tetrahedron with dihedral angles at all 6 edges equal to 2pi/v (such an object exists since this angle is smaller than pi/3). 
The number of tetrahedra you need for such a construction of course grows with k. For k=1, two tetrahedra suffice (see Thurston's knotted Y from his lecture notes).
A: Theorem(Long and Niblo): If $M$ is a 3-manifold and $S$ is an incompressible component of $\partial M$ then $\pi_1 S$ is separable in $\pi_1 M$ (pick a base point in $S$ to make sense of this).
This means that for any $\gamma\in \pi_1M\smallsetminus \pi_1 S$ there is a homomorphism $f$ from $\pi_1M$ to a finite group such that $f(\gamma)\notin f(\pi_1(S))$.
Corollary: If $|\pi_1M:\pi_1S|=\infty$ then $M$ had finite-sheeted coverings such that the preimage of $S$ has as many components as you like.
Proof: Choose a homomorphism $f$ such that $f(\pi_1S)$ has large index in $f(\pi_1M)$.  Now your covering corresponds to $\ker f$. QED 
So examples exists as finite covers of the example you already have!
