Ron Maimon's first two paragraphs of the initial answer was close to the truth (namely an emergence of a wall), but superselection is a red-herring.

(Disclaimer: it's not my own idea, but I've just learned this from the recent article arXiv:1606.05069.)

It's best to start from the Ising, a stat-mech. Let $F(h)$ be the free energy, and $C(M) = F(h) + hM$. $C$ is the free energy when $M$ is the variable. When $h = \pm 0$, $M = \pm M_{sp}$ and $C(+M_{sp}) = F(+0)$.

The key observation is that $C(M)$ is *not convex*. Its explicit expression is known for 2d n.n. ferromagnet (see the ref.) but we don't need it. Anyway $C(M) > C(M_{sp})$ when $|M| < M_{sp}$. More precisely $C(M) - C(M_{sp}) \propto L \to \infty$ in thermodynamic limit (TL). (Let the volume $V$ be $L^2$.) As Ron Maimon said, it's the cost of a domain wall.

Notice the last eq has to diverge. If you switch back the variable to $h$, $P(M) = \exp(-\beta C(M)) / Z(h)$, where $P(M)$ is the probability that the magnetization is $M$. In TL, it vanishes. (If you know the rudimentary of Monte Carlo simulation, it should be familiar.)

Yet the order of non-convexity of $C(M)$ inside the region $|M| < M_{sp}$ is $o(V)$. It's because $\partial C/\partial M = h$, er, it's lengthy and see the ref. What's important is that the inside region is $h = 0$, but we control the system by $M$. When there is first-order transition, an extensive variable is a good parameter. (When boiling water, throughout the entire "inside region" p remains 1atm and T = 100°C. Then the volume or energy specifies the state.)

The inside region is *not* a weird probabilistic mixture of $\pm M_{sp}$ equilibria. (Equilibirium in stat mech = vacuum in QFT.) Otherwise, SSB couldn't happen.

Are wall-states are equilibria? Yes, if you control $M$. No, if you control $h$, and $M$ is not constrained. In the outside region, both are valid, but in the inside region, the only available parameter is $M$. (For a real magnet, $M$ does not conserve, but you should already be familiar with such imaginary objects, like external source coupling to the field.)

The situation is the same for QFT. Intermediate states have a higher energy, but the true vacuum energy is not a "mexican hat." It has only the bump of $o(V)$. The "hat" states are metastable false vacua. (First order phase transition can have metastable states.)

Intermediate states are not a superposition of SSB vacua. $|+M_{sp}>$ and $|-M_{sp}>$ are residents of different theories, given in different Lagrangians $\mathcal{L}[\pm 0]$. They cannot be superposed. When the system is finite, any states can be added. But in TL, no.

The ref claims to prove translational invariance breaks for QFT. I don't know if states with different wall positions can be superposed when the boundary is periodic. Anyway when the boundary is free, translational invariance does not exist, and you see phonons in crystals, so I reckon it's not so much trouble even if translational invariance breaks. (Domain walls can only appear when the symmetry is discrete. For $O(N)$ models, do vortices appear?)

As for your original question...sorry my comments only add to confusions.

* Coleman-Weinberg model is an example where quantum correction breaks the classical symmetry, so your doubt in the original question has some point. (Good intro is in Peskin & Schroeder, last pages of the part 2.)

* You have to be careful distinguishing results and definitions, including the meaing of "near the classical vacuum"; when there's SSB, there seem to be many coupling constants, but they're related, and independent variables are fewer. Their relations are given by power expansion, and "near the classical vacuum" is the case where a weak-coupling theory is useful, I think.

* In QFT fields naturally have a dimension in power of length, but I'm not sure if their absolute value is meaningful; in classical electromagnetism, you have to introduce a new dimension to describe phenomena, unless you identify the charge element.

* I don't know well how to define coupling constants. For QED, static limit, i.e. Coulomb force can me measured. (The modern, precise way must be QHE or Josephson junction.) I've heard for weak force, it's defined by the tree-level interaction. (It's non-renormalizable.)