In systems that break continuous translation symmetry (regular crystals,lamellar solids, smectics etc.) one also necessarily breaks rotational invariance (but note that this is not true the other way around, allowing liquid crystals to exist). As you rightly pointed out the simple goldstone mode count does not seem to work in such ordered phases.
The reason is essentially because the orientational degrees of freedom get slaved to the translational phonon modes, which results in the orientational modes no longer being soft (they get gapped) and hence not Goldstone modes. This is analogous to the Anderson-Higgs mechanism but is different in that we only have global symmetries being spontaneously broken and no gauge fields involved in forming a crystal.
To see it explicitly, we can take for example a regular 3d crystal (with reciprocal lattice vectors ${\bf G}_n$). Then the relevant non-vanishing order parameter is going to be the fourier component of the mass density
$$
\rho({\bf r})=\rho_0+\sum_{{\bf G}_n}\left[\psi_{n}e^{i{\bf G}_n\cdot{\bf r}}+\mathrm{c.c}\right]
$$
where $\rho_0$ is the mean density and $\psi_n$ are the complex fourier amplitudes ($\langle|\psi_n|\rangle\neq0$ in the crystal). As the energy is invariant under uniform translations we require ${\bf r}\rightarrow{\bf r}+{\bf u}\implies\psi_n\rightarrow\psi_n e^{-i{\bf G}_n\cdot{\bf u}}$.
Now a global rigid rotation is also a symmetry of the energy: ${\bf G}_n\rightarrow{\bf G}_n+\delta\theta\times{\bf G}_n$. This leads to a corresponding displacement of lattice points as ${\bf u}=\delta\theta\times{\bf r}$.
As ${\bf u}$ is the translational goldstone mode and $\delta\theta$ is the rotational one, we immediately see that the orientational zero modes are slaved to the translation phonon modes ($\bf u$), which in this particular case is specified by
$$
\delta\theta_i=\dfrac{1}{4}\epsilon_{ijk}(\partial_j u_k-\partial_k u_j)
$$
which is the antisymmetric part of the strain tensor. As $u_i$ is a goldstone mode, its two point correlator will have a pole at zero wave-vector (momentum). This can also be obtained using linear classical elasticity for the phonons. Therefore heuristically (being cavalier about indices) as $\delta\theta\sim\partial u$ and $\langle|u(q)|^2\rangle\sim1/q^2$, we have $\langle|\delta\theta(q)|^2\rangle=$const. as $q\rightarrow0$ showing that the orientaional modes are no longer soft but are instead gapped (even though the rotational symmetry was spontaneously broken). This point was already noted by Mermin (Phys. Rev. 176, 250 (1968)) as it leads to true long-ranged orientational order even in 2d crystals.
You can also look at the book "Principles of Condensed Matter Physics" by Chaikin and Lubensky, where this point is discussed in multiple cases (including for the smectic-A phase, in which case a generalization of this point leads to the spectacular analogy between the SmA-Nematic transition to that of a superconductor).
Best Answer
You do have sound in $1+1d$. Mermin-Wagner do not claim that there are no massless particles in low dimensions -- indeed, some of the best understood QFTs are two-dimensional gapless theories, namely the wonderful world of $2d$ CFTs.
What Mermin-Wagner does, is proving that there are no order parameters for symmetry breaking in $2d$. This is not the same as saying there are no Goldstones. There are such particles, just no interpolating field for them.
See ref.1 for a nice recent review.
References: