I'm not entirely sure of this, comments appreciated
Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -> more stable while solvated -> better leaving group).
Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.
On the other hand, nucleophilic tendency is basically "how willing to sacrifice a lone pair is the group?". It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.
Another way to look at it would be via the "accessibility" of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to "give up" the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the reverse process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to "extract" it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more "accessible".
