This is a good question. I believe you are applying a mistaken analogy with an example that is often used to first introduce the notion of SSB: a single nonrelavistic particle in a double well with a potential barrier separating the two minima. As long as the barrier between the minima is finite, the particle can tunnel through it and inhabit both minima, so the reflectional symmetry is unbroken. But when the barrier height is formally taken to infinity, the particle gets "stuck" in one minimum or the other, breaking the symmetry. You are thinking that in a Mexican-hat-type potential, there is no potential barrier, so the system should be free to tunnel to all the minima, restoring the symmetry.
But an infinitely high potential barrier is a rather artificial notion. It's really a schematic representation of a more realistic situation in which, rather than a single particle, you have a spatially extended field (or a many-body system on a huge lattice) defined over a large volume $V$. Then the two minima in the one-particle picture really represent two distinct field configurations with equal total energy. Since going from one configuration to the other requires changing the value of the field at every single point in space (or spacetime), it requires a huge energy proportional to the total system volume $V$. So the "barrier height" in the one-particle picture really corresponds to the volume $V$ of the system in the field-theory picture. Only in the formal infinite-volume ("thermodynamic") limit is the symmetry truly broken.
Now moving the field configuration from, say, $\theta(x) \equiv 0$ to $\theta(x) \equiv \delta$ for some tiny angle $\delta$ only requires a tiny energy density proportional to $\delta$ (in the appropriate units), but the total energy $\delta \times V$ can still be very large. It's an order-of-limits subtlety: for any shift $\delta$ in the field value, no matter how small, we can image a system so large (roughly speaking, much larger than $1/\delta$) that shifting the entire system over by that amount requires an arbitrarily large amount of energy. So you still get an "infinitely high potential barrier to tunnel over" in the one-particle picture, even though the field's potential energy density $V(\varphi)$ has no barrier at all.
(To make things more explicitly quantum-mechanical, consider the quantum Heisenberg model on a lattice. If $|\psi\rangle$ and $|\psi'\rangle$ are two individual spin-$1/2$s rotated on the Bloch sphere by a small angle $\delta$, then the inner product $\langle \psi | \psi' \rangle = \cos \delta \approx 1 - (1/2) \delta^2$ is quite large, so a spin-$1/2$ could easily tunnel between the two states. But if we consider two huge systems of $N \gg 1$ aligned spins $|\Psi\rangle = \otimes_{i = 1}^N |\psi\rangle_i$ and $|\Psi'\rangle = \otimes_{i = 1}^N |\psi'\rangle_i$, then the tunneling amplitude $\langle \Psi | \Psi' \rangle = \cos(\delta)^n$ between the two systems is tiny, so it's very difficult to flip one state into the other.)
(To tie all this into TwoBs's answer: in the formal infinite-volume limit, the field's Fourier series representation becomes a continuous Fourier transform indexed by a continuous parameter $k$, and we can talk about Taylor expanding the energy density dispersion relation $\epsilon(k)$ about $k = 0$. For a Goldstone mode we have $\epsilon \to 0$ as $k \to 0$, but this is just an energy density - the total energy $E(k) = V \times \epsilon(k)$ is still huge, so the "energy barrier" is still very high. A Goldstone mode would need to be infinitely spatially extended in order to truly destroy the long-range order and restore the symmetry, and such an infinitely-large Goldstone mode is still too energetically costly to create.)