In some senses, protons exert an "electrostatic pressure" on each other. The bound neutron cannot become a proton, because of this electrostatic pressure applied by its companion protons - if it were to become a proton, it would need to get over the hurdle of that "pressure". In empty space, this "pressure" doesn't exist, so the neutron can easily slide into a proton-electron-antineutrino state.

As an analogy, when you're hot (say temperature t1) you take your jacket off. However, if you are in a really cramped room with a lot of other people, you'll need to become a lot warmer (say temperature t2) to go through the hassle of removing your jacket and elbowing your neighbour in the eye. If the room is air-conditioned so that you only ever get to temperature t1.5 (where t1 < t1.5 < t2), you'll not take your jacket off, even though you would if you were in a free state.

The basic idea is that a neutron all by itself has a lower energy state available to it, which is to decay into a proton, an electron, and an antineutrino. On the other hand, in the nucleus of a stable atom, a neutron decaying would result in a higher energy state for the whole nucleus.

Does that help?

The difference can be explained by potential energy. A neutron decaying in empty space causes a proton and an electron to wink into existence on level ground. A neutron decaying near a proton causes a proton to wink into existence on top of a cliff (the proton's electromagnetic field) and an electron to wink into existence at the bottom of a trench (ditto).

As far as what critical line is crossed, I can't really say. In fact, I'm not even sure if QCD / Electroweak Theory are yet able to calculate that.