Consider a simple 3-polytope P, (so every vertex has 3 neighbors). If $p_k$ is the number of faces of P which are k-gonal, Euler's theorem implies that $\sum_{k \ge 3} (6-p_k)=12$. Note that 6-gonal faces do not contribute to the LHS. (One way to think about it is that polygonal faces with 7 and more sides contributes "negative curvature", small faces contributes positive "curvature" and hexagons are "flat".)

Eberhard's theorem asserts that if you have a sequence of numbers $p_k, k \ne 6$ such that $\sum_{k \ge 3} (6-p_k)=12$ then you can find a simple 3-polytope with $p_k$ k-gonal faces. (But you have no control on $p_6$). 


There are various results extending Eberhard's theorem in various directions. Chapter 13 in Grunbaum's book "[Convex Polytopes][1]" and esspecially the supplemantary material at the end of the chapter in the new edition is a good source. Another general source is [my chapter][2] from the "Handbook of Discrete and Computational geometry" on garphs and skeleta of polytops.  


A relatively recent paper on the subject is by Stanislav Jendrol "[On the face vectors of trivalent convex polyhedra][3]". Another paper by Jendrol which deals with general 3-polytopes from the same year is "On face vectors and vertex vectors of convex polyhedra" Discrete Math 118 (1993)119-144. There are analogs of Eberhard theorem for 4-regular planar graphs, for toroidal graphs and in other directions.

In higher dimensions and even in four dimensions these problems become very difficult and very little is known? You can ask again about the numbers of k-gonal 2-dimensional faces. While the formula above implies that in dimension 3 and more $p_3+p_4+p_5>0$ it is known that in dimensions 5 and more $p_3+p_4>0$. You can also ask for a simple 4 polytope (a 4-polytope whose graph is 4-regular) what are the number of facets $p_Q$ isomorphic to a given 3-polytope Q. This gives you a vector indexed by combinatorial types of simple 3-polytopes, but I am not aware of any Eberhard type theorem and I do not know even which 3-polytopes should be considered as the analogs of the hexagons in the above formula.


  [1]: http://books.google.co.il/books?id=ISHO86XJ1CsC&printsec=frontcover&dq=grunbaum's+convex+polytopes&source=bl&ots=S7T41O6hvF&sig=wRkbvA0TtPVeRuiRFHo6FjVQv1I&hl=iw&ei=QkrqS7LzCaOiOLaTsPMK&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAgQ6AEwAA#v=onepage&q=grunbaum's%2520convex%2520polytopes&f=false
  [2]: http://www.ma.huji.ac.il/~kalai/ch19.pdf
  [3]: http://pdf.dml.cz/bitstream/handle/10338.dmlcz/136327/MathSlov_33-1983-2_7.pdf