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This is a variation of Agol's answer. Take any contractible special polyhedron $X$, that is a compact polyhedron such that every point has a neighborhood of one of these three types:

and such that the natural stratification is a cellularization, i.e. all the points as in the center (right) form disjoint open 1-cells (2-cells). A famous example is Bing's house:

Now thicken 4-dimensionally the polyhedron $X$. You do this first by thickening the 1-skeleton, and this can be done in a unique way (we only construct orientable 4-manifolds). The rest can be thickened in various ways, but for simplicity we restrict ourselves only to locally flat thickenings, i.e. thickenings that locally lie in a (smooth) 3-dimensional slice (a very reasonable assumption). As shown by Turaev, the locally flat thickenings are parametrized by assigning a (half-)integer at each 2-cell of the polyhedron (these numbers are called gleams), which measures the "Euler number" of the thickening. This is very similar to the description of the 4-manifold via a Kirby diagram. For instance in the Bing's house there are three 2-cells and each needs to be coloured by an integer.

The boundary 3-manifolds of all these thickenings are all obtained by Dehn surgery on a fixed link $L$ in a connected sum $\sharp_h(S^2\times S^1)$ of some $h$ copies of $S^2\times S^1$. Different gleams yield different surgeries on the same link $L$. The nice thing here is the following: Costantino and Thurston have shown that the link $L$ is always hyperbolic; the complement $\sharp_h(S^2\times S^1)\setminus L$ can be constructed by gluing some regular ideal octahedra. You need two regular octahedra for each vertex of $X$, so that the volume of the hyperbolic manifold is precisely twice the volume of the regular ideal hyperbolic octahedron times the number of vertices of $X$.

Now as Agol said you only need to use Thurston Dehn filling theorem that ensures you that by avoiding a finite set of gleams on each face the resulting 3-manifold is hyperbolic. That is, the boundary of the resulting thickened 4-manifold is hyperbolic. Moreover the 6 theorem tells you that if all the gleams are bigger than 6 you are certainly done (because we know very well how the cusps are). Hence you can decorate Bing's house with a triple of integers bigger than 6 and you get plenty of hyperbolic 3-manifolds bounding contractible 4-manifolds.

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This is a variation of Agol's answer. Take any contractible special polyhedron $X$, that is a compact polyhedron such that every point has a neighborhood of one of these three types:

and such that the natural stratification is a cellularization, i.e. all the points as in the center (right) form disjoint open 1-cells (2-cells). A famous example is Bing's house:

Now thicken 4-dimensionally the polyhedron $X$. You do this first by thickening the 1-skeleton, and this can be done in a unique way (we only construct orientable 4-manifolds). The rest can be thickened in various ways, but for simplicity we restrict ourselves only to locally flat thickenings, i.e. thickenings that locally lie in a (smooth) 3-dimensional slice (a very reasonable assumption). As shown by Turaev, the locally flat thickenings are parametrized by assigning a (half-)integer at each 2-cell of the polyhedron (these numbers are called gleams), which measures the "Euler number" of the thickening. This is very similar to the description of the 4-manifold via a Kirby diagram. For instance in the Bing's house there are three 2-cells and each needs to be coloured by an integer.

The boundary 3-manifolds of all these thickenings are all obtained by Dehn surgery on a fixed link $L$ in a connected sum $\sharp_h(S^2\times S^1)$ of some $h$ copies of $S^2\times S^1$. Different gleams yield different surgeries on the same link $L$. The nice thing here is the following: Costantino and Thurston have shown that the link $L$ is always hyperbolic; the complement $\sharp_h(S^2\times S^1)\setminus L$ can be constructed by gluing some regular ideal octahedra. You need two regular octahedra for each vertex of $X$, so that the volume of the hyperbolic manifold is precisely twice the volume of the regular ideal hyperbolic octahedron times the number of vertices of $X$.

Now as Agol said you only need to use Thurston Dehn filling theorem that ensures you that by avoiding a finite set of gleams on each face the resulting 3-manifold is hyperbolic. That is, the boundary of the resulting thickened 4-manifold is hyperbolic. Moreover the 6 theorem tells you that if all the gleams are bigger than 6 you are certainly done (because we know very well how the cusps are). Hence you can decorate Bing's house with a triple of integers bigger than 6 and you get plenty of hyperbolic 3-manifolds bounding contractible 4-manifolds.