I didn’t want you to infer anything from my earlier comment about acid-base equilibrium. I thought I had clearly stated that your comment about the concentration of hydroxide concentration being “low” was most likely to be erroneous given the question, as one uses indicators in extremely small amounts. If that was unclear, I should have been more emphatic about the point. If one is examining the behavior of an acid with a base (such as sodium hydroxide), one adds a very small amount of an indicator so that the acid-base chemistry is predominantly due to the acid and base under investigation, not the indicator. If one is working with 10 milliMolar base, then one will probably only need an indicator at, say, 10 microMolar concentration, or thereabouts.
Insofar as the reaction mechanism – certainly, while nucleophilic additions to simple, unpolarized alkenes is generally not seen to my knowledge (simple nucleophilic additions are typically to carbonyl carbons, given the polarity of the C=O double bond), when you’ve got a large molecule like phenolphthalein that is resonance-stabilized and asymmetric about that central carbon (two phenol substituents and a benzoate substituent), the observation of a nucleophilic addition to its central sp2 hybridized carbon does not surprise me. After all, it is hardly the same as a simple, unsubstituted alkene. And yes, breaking up the conjugated structure across the two phenol substitutents is certainly why the color fades under strongly basic conditions, barring some unforeseen subtlety I haven’t considered.
Sure, it’s certainly possible that hydroxide ions can and do add to phenolphthalein, generating that colorless structure, under less basic conditions. We are dealing with a bulk sample that is macroscopically observable, after all. But it would also be logical that it would not be very many, and that it is not enough to change the color very much given the state of the rest of the system. It is also likely that such a structure would be disfavored for whatever reason, and would find its way back to the pink structure.
If you feel the need to further pursue this inquiry, I am afraid I won’t be of much help.