How to calculate the likely size of an OLAP cube
I can see some roadblocks to creating this estimate. Knowing the row counts and cardinalities of the dimension tables in isolation isn't nearly as important as the relationships between them.
Imagine two low-cardinality dimensions with n and m unique values respectively. Caching OLAP aggregates over those dimensions produces anywhere from n + m values to n * m values depending on how closely the relationship resembles a pure bijection. Given only the information you provided, all you can say is you'll end up with fewer than 3.64 * 10^34 values, which is not very useful.
I'm pessimistic there's an algorithm fast enough that it wouldn't make more sense to generate the cube and weigh it when you're done.
We wrote a research paper that seems relevant:
Kamel Aouiche and Daniel Lemire, A Comparison of Five Probabilistic View-Size Estimation Techniques in OLAP, DOLAP 2007, pp. 17-24, 2007. http://arxiv.org/abs/cs.DB/0703058
Well. You can use a general rule of Analysis Services data being about 1/4 - 1/3 size of the same data stored in relational database.
Edward.