My general question is: is it consistent that any uncountable family of less than $\mathrm{non}(\mathcal{N})$ sets, each with positive outer measure, has an uncountable subfamily $\mathcal{F}$ s.t. $\bigcap \mathcal{F}$ has positive outer measure?

Here, $\mathrm{non}(\mathcal{N})$ denotes the uniformity number of the ideal of null sets, i.e. the smallest cardinality of a set that is not of measure zero.

The original question I was thinking about was:

Is it true that every uncountable family of sets of positive measure has an uncountable subfamily with an intersection of positive measure?

It is not difficult to see that under CH, the answer is strongly negative, namely if $\mathfrak{c} = \omega_1$, then there exists an uncountable family $\mathcal{X}$ of sets of even full measure such that every uncountable subfamily of $\mathcal{X}$ has even an empty intersection. To see this, order the reals in an $\omega_1$-sequence $\{x_\alpha\}_{\alpha < \omega_1}$ and for each $\alpha < \omega_1$ define a tail of the reals above $x_\alpha$, i.e. let $X_\alpha = \{x_\beta: \alpha < \beta < \omega_1\}$. Then, the complement of each $X_\alpha$ is a null set, so all the sets in $\mathcal{X} = \{X_\alpha\}_{\alpha < \omega_1}$ are of full measure, yet every real $x$ belongs to at most countably many sets $X_\alpha$, so it cannot be a member of any intersection of uncountably many elements of $\mathcal{X}$.

Now, observe that the same reasoning seems to work in any model of ZFC, where $\mathrm{non}(\mathcal{N}) = \aleph_1$. If this is the case, then, by definition there exists a set $A \subseteq \mathbb{R}$ with $\mu^\ast(A) > 0$ and $|A| = \aleph_1$ (of course, if CH does not hold, $A$ cannot be measurable, but it does not matter here). Apply the same reasoning to $A$ as we applied above to the set reals.

Furthermore, this actually shows that there always exists a family of size $\mathrm{non}(\mathcal{N})$ sets of positive outer measure such that any intersection of at least $\mathrm{non}(\mathcal{N})$ many elements of this family is empty. We just take $A \subseteq \mathbb{R}$ of size $\mathrm{non}(\mathcal{N})$ and $\mu^\ast(A) >0$, and repeat the argument with $A = \{x_\alpha: \alpha < \mathrm{non}(\mathcal{N})\}$, and the sets $X_\alpha = \{x_\beta: \alpha < \beta < \mathrm{non}(\mathcal{N})\}$, where $\alpha < \mathrm{non}(\mathcal{N})$ - for each element $x$ of $A$ the number of sets $X_\alpha$ with $x \in X_\alpha$ is bounded in $\mathrm{non}(\mathcal{N})$.

So, it is true that there is always a family of at least $\mathrm{non}(\mathcal{N})$ sets of positive outer measure such that each intersection of $\mathrm{non}(\mathcal{N})$ of its elements is actually empty.

Two questions, then:

(1) Is it consistent that each uncountable family of strictly less than $\mathrm{non}(\mathcal{N})$ sets of positive (outer) measure has an uncountable subfamily (a) of same size, or (b) smaller, yet uncountable, with an intersection of positive measure?

(2) Is it consistent that each uncountable family at least $\mathrm{non}(\mathcal{N})$ sets of positive (outer) measure has a strictly smaller, yet uncountable subfamily with an intersection of positive measure?

An additional, loosely related, question out of curiosity - it is known that if we take any family of less than $\mathrm{add}(\mathcal{N})$ (which denotes the additivity number, i.e. the minimal cardinality of null sets such that their sum is not null) sets of full measure, then their intersection also has the full measure. It is obvious that this statement does not hold if "full" is replaced with "positive (outer)" in the above - simply take the sets $[k, k+\epsilon]$ for some fixed small $\epsilon$, and for all integers $k$ - the intersection of this countable family is empty. But what about uncountable collections of strictly less than $ \mathrm{add}(\mathcal{N})$ sets of positive (outer) measure: can there be such families with null intersection?

And even more loosely related question - it is obvious that e.g. the uniformity number would not make much sense in the context in which we restricted the attention to measurable sets of reals (namely, each measurable set of positive Lebesgue measure has to have cardinality continuum). But, it is not obvious to me if the same holds for the additivity number - if we restrict attention to measurable sets of reals, would it also hold that this measurable-additivity would be equal to continuum?