First of all, not only the existence of a free ultrafilter on $\omega$ is far weaker than $\sf BPI$, even the statement that every filter on $\omega$ can be extended to an ultrafilter on $\omega$ is strictly stronger than "There is a free ultrafilter on $\omega$". This can be an example for a principle that you're looking for: Every filter on $\omega$ can be extended to an ultrafilter. It is weaker than $\sf BPI$, and it proves the existence of a free ultrafilter on $\omega$.

Secondly, there is no "formal definition" of what are choice principles. See What is a Choice Principle, really? for two possible answers, which do not agree at all on the meaning of a choice principle.

Finally, there is a notion of strength when it comes to statement provable from $\sf ZFC$ and not from $\sf ZF$ (or even just "consistent with $\sf ZFC$", like $V=L$ or $\sf GCH$ which imply choice but are not equivalent to it). This is the question whether or not one statement implies the other. So $\sf BPI$ implies $\sf BPI(\omega)$ which implies the existence of a free ultrafilter on $\omega$, and none of these can be reversed. So we have a good notion of being stronger: a stronger principle proves more.

Let me finish by saying that "Every countable family of non-empty sets admits a choice function" cannot prove, nor it is a consequence of $\sf BPI$, so if by a choice principle you mean something like "Every such and such family of sets admits a choice function", then in all likelihood you're expecting a failure; unless you allow something like "There is a choice function from every family which is used in the proof that a filter can be extended to an ultrafilter", or something like that.

First of all, not only the existence of a free ultrafilter on $\omega$ is far weaker than $\sf BPI$, even the statement that every filter on $\omega$ can be extended to an ultrafilter on $\omega$ is strictly stronger than "There is a free ultrafilter on $\omega$". This can be an example for a principle that you're looking for: Every filter on $\omega$ can be extended to an ultrafilter. It is weaker than $\sf BPI$, and it proves the existence of a free ultrafilter on $\omega$.

Secondly, there is no "formal definition" of what are choice principles. See What is a Choice Principle, really? for two possible answers, which do not agree at all on the meaning of a choice principle.

Finally, there is a notion of strength when it comes to statement provable from $\sf ZFC$ and not from $\sf ZF$ (or even just "consistent with $\sf ZFC$", like $V=L$ or $\sf GCH$ which imply choice but are not equivalent to it). This is the question whether or not one statement implies the other. So $\sf BPI$ implies $\sf BPI(\omega)$ which implies the existence of a free ultrafilter on $\omega$, and none of these can be reversed. So we have a good notion of being stronger: a stronger principle proves more.

Let me finish by saying that "Every countable family of non-empty sets admits a choice function" cannot prove, nor it is a consequence of $\sf BPI$, so if by a choice principle you mean something like "Every such and such family of sets admits a choice function", then in all likelihood you're expecting a failure; unless you allow something like "There is a choice function from every family which is used in the proof that a filter can be extended to an ultrafilter", or something like that.