It is a consequence of  Malgrange's preparation theorem for differentiable
functions  that $C^{\infty}(M)$ is a faithfully flat  $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of
analytic functions on $M$). See Corollary 1.12, Chapter VI of
his book "Ideals of differentiable functions".

On the other hand  $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out
by Greg Stevenson shows.

I believe that these two facts can be putted together to give a positive answer to
the question.