It does not coincide with projective dimension because a projective module is not necessarily isomorphic to a free module (while a projective $P$ has a trivial projective resolution $0 \to P \to P \to 0$). In fact, projective modules that admit a free resolution are precisely the ones that are stably free (i.e. become free when a free module is added). It is known that this is always the case for finitely generated modules over a polynomial ring (this is a theorem of Serre). Cf. the last chapter of Lang's Algebra.
It is, incidentally, still true that the free dimension is finite in many interesting cases, such as finitely generated modules over a regular local ring or a polynomial ring over a field. This is because, in the former case, the residue class field $k$ of a regular local ring $R$ has a finite free resolution (namely, the Koszul complex), and any f.g. module $M$ with $\mathrm{Tor}_R(M,k) = 0$ can be shown to be projective, hence free (because $R$ is local, by a Nakayama argument). (Added- admittedly in the local case this follows from the fact that projective dimension is finite. In the graded (polynomial) case a projective module is still free, but this is at least harder to prove.)
Edit: As pointed out in the comments, "free dimension" should require finitely generatedness of the resolution.