Why don't low frequency phonons scatter electrons in a superconductor?

Typical phonon energies are in the meV range. Typical Cooper pair binding energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a Cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{rm B}T approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{rm B}cdot4.2{rm K}approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.

Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.

To be in the superconducting state, the temperature must be below the superconducting gap $2Delta$ (modulo constants of order 1).

This automatically means that thermal energy $k_B T$ is lower than the superconducting gap. Thus, the only thermally populated excitations, be they phonons or anything else, do not have enough energy to break a Cooper pair.

But what if we don't break a Cooper pair, but just change its total momentum slightly? This is definitely possible through scattering via acoustic phonons for example, which can have very low momentum. However, the issue here is Pauli exclusion, all the other momentum states for the Cooper pair are occupied by other Cooper pairs. Because Cooper pairs are built from two fermions, Pauli exclusion still holds even though they are composite bosons. So small angle scattering is also suppressed.