Why does using NaI as a nucleophile and acetone as solvent lead to an SN2 reaction mechanism instead of SN1? Is I- not considered a weak nucleophile?
As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:

$$\ce{NaI + RX -&gt; NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$ 

This is in acetone solvent (I don't know how to format it along the arrow).

If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.

While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$. 

This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.