This is not a well-defined question, as there is not necessarily a notion of "regular polyhedron". You might require that all faces are equivalent; or all edges are equivalent; or all vertices; or some combinations of these. These will give you sets of Platonic solids, Archimedean solids, Catalan solids, prisms, bipyramids, antiprisms, or trapezohedra, depending which conditions you impose. But in general, there is not necessarily any particularly symmetric polyhedron with a given number of vertices (or faces, or edges). The first 3 sets of solids above are highly symmetric and probably similar to what you want, but are finite in size, and highly irregular: you would need to hard-code every formula, essentially. The other 4 sets will generalize to arbitrary $n$ (or $2n$, or so) but will likely not have symmetries that you want. Their coordinates are given by very similar formulas to what you have for 2D.
In case this doesn't make sense, why your question is ill-posed, please consider: what kind of answer would you want from your formula for the 5-vertex case, radius 1?