Any N-particle system in 3 dimensions has 3N degrees of freedom. 3 are pure translations and 3 are pure rotations; hence the typical formula for molecules as having 3N-6 internal (vibrational) degrees of freedom.
For a macroscopic solid, N is huge (a solid weighing a few grams will have nearly Avogadro’s number of atoms), so the internal degrees of freedom dwarf the translations and rotations. But of course a finite object can still rotate and translate—it’s just that those motions don’t contribute significantly to the quantum description of the solid.
In the idealized case of an infinite crystal (like you might encounter in a class on solid state physics), nuclear motions are generally only considered over a single unit cell. In this case, if the unit cell contains N atoms, it still has 3N degrees of freedom (assuming the cell is 3-dimensional). But now the atomic motions of a unit cell are going to affect the motions of the neighboring cells. 3 of the modes will look like translational modes—these correspond to acoustic phonons and have zero energy at zero crystal momentum. The other 3N-3 modes correspond to optical phonons, which have a nonzero energy at zero crystal momentum.
Edit for clarity: in general, any atom in a crystal can move in any direction. However, these motions can be decomposed into a set of 3N normal modes—eigenstates of the vibrational Hamiltonian that are orthogonal to one another (that is, their ##L^2## inner product is zero).