[I answered a bit to quickly, maybe that's not what you are after ?]There is something in $Cat/B$ : $Alg(T) \to B$ is the universal monadic right adjoint through which $F$ factors.

There's probably a cleaner way to see this, but here is how I think about it :

Indeed, it is easy to see that a factorisation of $F :A \to B$ through a monadic functor $B^M \to B$ is the same as an action of the monad $M$ on $F$.

Now, codensity monad are the same "endomorphisms monad" (in the sense of endomorphism objects in a monoidal category acting on a category), so such an action is the same as a morphism of monad $M \to End(F)$ which in turn corresponds to a forgetfull functor $B^{End(F)} \to B^M$ over $B$.

Putting together, you get that a functor $A \to B^M$ over $B$, is the same as functor $B^{\tilde{F}} \to B^M$ over $B$.

There is something in $Cat/B$ : $Alg(T) \to B$ is the universal monadic right adjoint through which $F$ factors.

There's probably a cleaner way to see this, but here is how I think about it :

It is easy to see that a factorisation of $F :A \to B$ through a monadic functor $B^M \to B$ is the same as an action of the monad $M$ on $F$.

Now, codensity monad are the same as "endomorphisms monad" (in the sense of endomorphism objects in a monoidal category acting on a category), so such an action is the same as a morphism of monad $M \to End(F)=T$ which in turn corresponds to a forgetfull functor $B^{End(F)} \to B^M$ over $B$.

Putting together, you get that a functor $A \to B^M$ over $B$, is the same as functor $B^{End(F)} \to B^M$ over $B$.

I guess an advantages of this point of view is that it is easier to generalizes $(\infty,1)$-categories as the endomorphism monad point of view is how Lurie deals with monadicity in Higher Algebra...