power series ring is faithfully flat but not free

If $A =\mathbb Z$, then as a $\mathbb Z$-module, there is an isomorphism $$\mathbb Z [[x]] \cong \prod_{i = 0}^{\infty} \mathbb Z.$$ The fact that this abelian group is not free is a standard fact, discussed e.g. here (indirectly; if it were free, its $\mathbb Z$-dual would be huge, rather than just $\bigoplus_i \mathbb Z$; see here) and here.

Note that if e.g. $A$ is a field then $A[[x]]$ is free over $A$, so this depends on the ring $A$.

One way to see flatness when $A$ is Noetherian is that $A[x]$ is obviously flat over $A$ (being free), and completions of Noetherian rings are flat over the original ring; hence $A[[x]]$ is flat over $A[x]$, and so over $A$. For faithful flatness, one can then use that there is a surjection $A[[x]] \to A$ given by mapping $x \to 0.$

In general (i.e. the non-Noetherian case), one has that for a finitely presented $A$-module, there is a natural isomorphism $M\otimes_A A[[x]] \cong M[[x]]$, so that $\text{--}\otimes_A A[[x]]$ is exact on finitely presented $A$-modules.

If $A$ is furthermore coherent (e.g. Noetherian!) then any finitely generated submodule of a finitely presented module is itself finitely presented, and so we may write an injection of arbitrary $A$-modules as the direct limit of an injection of finitely presented $A$-modules. Since tensor product commutes with direct limits, we conclude (when $A$ is coherent) that $\text{--}\otimes_A A[[x]]$ is exact on arbitrary $A$-modules, and hence that $A[[x]]$ is $A$-flat. Faithful flatness again follows from the existence of the map $A[[x]] \to A$. There are rings that are coherent but not Noetherian (see e.g. here, so this is slightly more general.

Actually, as is explained here, if an arbitrary product of flat $A$-modules is again flat, then $A$ is necessarily coherent. I'm not sure what the story is for the countable product $A[[x]]$ of copies of $A$ (i.e. exactly what condition is necessary for this particular $A$-module to be flat), but as noted in the link, if $A = k[t_1,t_2,\ldots]/(t_i t_j)_{i,j = 1,\ldots,\infty},$ then $A[[x]]$ is not flat over $A$.

(Thanks to user26857 for their comment below, which led me to correct a blunder in an earlier version of this answer, and drew my attention to the above link.)