As WimC points out, there is no difficulty extending the concept of a Delaunay triangulation to higher dimensions. In fact, Delaunay himself worked with a periodic point set in a Euclidean space of arbitrary finite dimension (you can now find his paper online: "Sur la sphère vide").
It is true that you cannot tile $\mathbb{E}^3$ with regular tetrahedra, but there is no requirement that the Delaunay simplices be regular. There is a technicality, that is typically not given the attention it deserves: not every set of points will yield a Delaunay triangulation. For example, if you choose the vertices to be the integer lattice (cartesian lattice), you encounter a problem: the points are not in "general position".
The Delaunay complex on a set of points $P \subset \mathbb{E}^n$ can be characterised as the set of simplices that have an open circumscribing ball that does not intersect $P$. If the points are not in general position you obtain simplices of dimension higher than $n$, and therefore the complex is not embedded; it is not a triangulation. As Delaunay pointed out, this only happens when you have more than $n+1$ points on some sphere, and such configurations can be avoided with an arbitrarily small perturbation. This is why the problem is often ignored, by assuming that $P$ is in general position.
Anyway, once you are aware of that issue, you can get a nice Delaunay triangulation in $\mathbb{E}^n$ by using your favourite lattice. For example, the Delaunay tetrahedra in the BCC lattice are all mutually isometric.
The Delaunay complex can be defined in the same way when $P \subset M$, for an arbitrary metric space $M$. If $M$ is a manifold with constant curvature, then you will get a triangulation, provided the vertices are sufficiently dense, and you avoid degenerate configurations. However, if the curvature is not constant, you need to work harder to obtain a triangulation: you cannot have $P$ "arbitrarily close" to a degenerate configuration.
