Does this activation of carbonyls usually involve a SN1 or a SN2?
It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction.  It is commonly referred to as an "addition-elimination" reaction.  You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate.  Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction.  This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl.  Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously.  Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product.  You "add" a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you "eliminate" one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.



Here's another, more specific drawing for the case at hand.