The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the outer density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive measure, then we can split $U$ in half $U=A\sqcup B$ each with half the outer density. One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the measure of $U$. By iterating this, we can make the outer measure of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.