It’s become clear that this answer is wrong. I’m leaving it here to preserve valuable discussion in the comments until someone formulates a correct answer.
Consider the power set in Set. If we have a function f : S -> T, we can form f' : PS S -> PS T by f' X = f [X]. Nice covariant functor (I think). We could also form f'' X = f^(-1) [X], a nice contravariant functor (I think).
Let’s look at the “power set” in Haskell:
newtype PS t = PS (t -> Bool)
This is not a Functor, but only a Contravariant:
instance Contravariant PS where
contramap f (PS g) = PS (g . f)
We recognize this because t is in negative position. Unlike Set, we cannot get at the “elements” of the characteristic functions that make up the power set, so the covariant functor isn’t available.
I would conjecture, therefore, that the reason Haskell admits a free monad for every covariant functor is that it excludes those covariant functors that cause trouble for Set.