How "strong" is the existence of a non trivial ultrafilter on $\omega$? Obviously the question in the title alone doesn't make sense so I'll develop on the context and then I'll ask my question :
Studying $AD$ (axiom of determinacy) I had to prove that $AD$ and $AC$ are incompatible (mod $ZF$). So to prove this I show that under $AC$, there are some undetermined games, and in order to prove this I use the fact that the Fréchet filter can be extended to an ultrafilter (necessarily non principal). As a remark, I thought "This shows that $AD$ is incompatible, not only with $AC$, but also with strictly weaker choice principles, such as the $BPI$ or the statement 'There exists a non-principal ultrafilter over $\omega$' ". Thinking this I wondered how strong that last statement (let $\Omega$ denote said last statement) was, and whether it could imply $BPI$. The answer seems to be "Obviously, no", as $\Omega$ is only about a specific set ($\omega$), whereas the $BPI$ is much more general. But then I wondered whether there was any "general choice principle", strong enough to prove $\Omega$, but not strong enough to prove the $BPI$. I then looked for a way to make "general choice principle" precise in order to look for an answer. So here are my questions :

Is there a satisfactory way to make the notion of "general choice principle" precise, i.e. characterize certain sentences such that $AC$, $BPI$, etc. fall under this scope, but $\Omega$ doesn't ? 
If there is, let $\phi$ be such a general choice principle. Can we have $ZF + \phi \vdash \Omega$, but $ZF + \phi$ doesn't prove $BPI$ ?

I first thought of defining it as "$ZF^{-}+ \phi$ ($ZF^{-}$ being $ZF$ minus the axiom of infinity) does not prove the axiom of infinity", but 1.it was only for this particular example, 2. it didn't work, as "$Inf \implies \Omega$" would fall under this notion, but clearly wouldn't be satisfcatory, so a solution would have to be more clever than that. 
 A: It seems difficult to make the notion of "general choice principle" precise, but I would guess that the principle "every infinite set admits a nonprincipal ultrafilter" would qualify.  If so, then it answers your second question. It obviously implies the existence of a nonprincipal ultrafilter on $\omega$, and, unless I'm making a stupid mistake, it doesn't imply BPI because it follows from the conjunction of "there is a nonprincipal ultrafilter on $\omega$" and countable choice.  (The point is that countable choice implies that there are no infinite, Dedekind-finite sets, so every infinite set includes a copy of $\omega$.)
A: 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.
