Applying the logic found here would suggest the opposite. This is because both conjugate bases delocalize their negative charge, but while ethanoic acid has two oxygens to delocalize its charge among, phthalimide has two oxygens and a nitrogen (and a benzene ring); it seems that the quality of the delocalization is the same, but the quantity favors phthalimide (as the answers to the question put it). As for the other factors that affect acidity (inductive effects, hybridization, atom size/electronegativity, formal charge, etc.) none seem to come into play.
The deprotonated phthalimide anion is indeed highly delocalized, but the neutral molecule with the N-H proton reattached retains most of that delocalization. Protonating acetate ion back to acetic acid causes a much greater loss of delocalization.