The nitrogen is connected to a C on either side, bonded to one 1, and has 1 lone pair, right? So it's not really susceptible to nucleophilic attack it seems, since it's plenty electron-sufficient.
I think the H+ will protonate the N, producing a 4 coordinate N (with a formal + charge) transitionally. The electrons in the N-C bond will be pushed to N, and the OH- simultaneously will attack the C.
Is the methyl on a carbon next to the N or across the ring? if it's across the ring it doesn't matter, but if it's next to the ring, you have to decide which N-C bond will be broken, in which case I'd say the one to the carbon with the Methyl attached, since the methyl would stabilize the formal negative transition state through hyperconjugation...
Long story short, I think the product is:
NH2-CH2-CH2-CHMeOH if the Me is on the carbon next to N
or
NH2-CH2-CHMe-CH2OH if the Me is on the carbon across from N