Fibers of a generic smooth map are polyhedra (so in particular CW-complexes) by the triangulation conjecture of Thom, proved by Andrei Verona [Stratified Mappings - Structure and Triangulability, Springer LNM vol. 1102]. I'm not sure that the full strength of the triangulation conjecture is really needed here - hopefully Ryan's projected answer will clarify this. A more general triangulation conjecture was proved by Masahiro Shiota [Thom’s conjecture on triangulations of maps, Topology, 39 (2000), 383–399], who also has further results on this subject on the arXiv.
Verona's theorem says that any proper, topologically stable smooth map between smooth manifolds $M$ and $N$ is "triangulable", i.e. equivalent to a PL map by a topological change of coordinates in $M$ and in $N$. On the other hand the set of proper topologically stable smooth maps $M\to N$ is dense in $C^\infty(M,N)$ by the Thom-Mather theorem.
So don't worry, there are no Hawiian Hawaiian earrings or other nightmares hiding in the fibers. Since To ensure that nothing obstructs easy sleep, we want the fibers of a generic smooth map $M^m\to N^n$ have to be polyhedra of dimension $\le\max(m-n,0)$, \le\max(m-n,0)$. Indeed, forgetting about the "polyhedra" part for a moment, we know the dimension estimate from multijet transversality (as sketched by Tom). We can now achieve "polyhedra" and "of dimension $\le\max(m-n,0)$" to hold simultaneously because maps generic in two ways are still generic. (in That is, the sense intersection of an two open and dense subset subsets of $C^\infty$), C^\infty(M,N)$ is open and dense. I guess this needs $M$ to be compact, and the fibers of a generic smooth map general case follows by writing $M\to N$ are polyhedra M$ as a union of dimension $\le\max(m-n,0)$.an increasing chain of compact submanifolds and applying Baire's theorem.)
