It sort of depends on what exactly you mean by "settled". Up to physics standards, the question is definitely settled: it has been understood, for many decades now, that $\phi^4$ in $d=4$ is trivial. The only open question is to prove this in a way that mathematicians will find convincing. This is very similar to the Yang-Mills mass gap problem: physicists have known that YM is gapped for many decades now, it only remains to prove it in a mathematically rigorous way. But the answer is definitely "known".
We actually know the answer in any dimension. In $4d$, the only non-trivial theory is non-abelian Yang-Mills with a handful of bosons and fermions. If you add too many matter fields, or make the gauge field abelian, you make the theory trivial. In lower dimensions, any gauge theory is non-trivial, but e.g. $3d$ $\phi^6$ is again trivial. In $2d$, any $V(\phi)$ is non-trivial.
This is the correct, known answer. It is not "settled" only in the sense that we don't know how to prove this rigorously, but only because we don't have a rigorous definition of QFT. As soon as we manage to define QFT in a rigorous way, one should be able to prove these claims rigorously; absolutely nobody expects a surprise here: if your rigorous definition of QFT does not imply triviality of $\phi^4$, or a YM gap, then your definition is wrong.
It is important to stress that "triviality" in this context does not mean that the theory is boring or useless. For example, $4d$ QED is trivial, but it is obviously a tremendously useful theory, it makes the most accurate physical prediction ever. In this context, "triviality" simply means "infrared triviality", i.e., the coupling constant becomes smaller and smaller the lower the energy of your experiment is. QED and $\phi^4$, and many other theories, are infrared trivial, but they are still non-trivial, interacting, useful low-energy effective theories.
