For some integer $n$, consider $S=\{1, \dots, n\}$. Let $S_1, \dots, S_k$ be subsets that satisfy the property that each $x\in S$ is in exactly 3 of the $S_i$'s and so that the intersection of these three subsets is $x$. What is the minimum value of $k$ (the number of subsets)?
I thought that if each $x$ was in 2 of the sets, $k$ would be at least $2*\sqrt{n}$, but was not sure how to proceed for the case of 3 sets.
I have updated this answer with a purely combinatorial description, since the original problem was combinatorial in flavor.
Choose $k$ so as small as possible so that $\binom{k}3\ge n$. Identify the numbers $\{1,\dots,\binom{k}3\}$ with the subsets of $\{1,\dots,k\}$ of size $3$. Then, for each $i\in\{1,\dots,k\}$, let $S_i$ be the set of numbers in $\{1,\dots,n\}$ whose corresponding subset contains $i$. Clearly, any $x\in\{1,\dots,n\}$, corresponding to the set $\{i,j,h\}$, is contained in exactly three subsets $S_i,S_j,S_h$, and the intersection of these subsets is $x$.
This is optimal for the reason lulu said in a comment; in order for such a collection of sets to exist, you you need $\binom{k}3\ge n$. Asymptotically, we get $k\sim \sqrt[3]{6n}$.
This generalizes to any multiplicity of intersections. In particular, $2\sqrt{n}$ for the two-way intersection problem you mentioned is not optimal, and you can obtain $\sim\sqrt{2n}$ by letting $\{1,\dots,\binom{k}2\}$ correspond to pairs of numbers in $\{1,\dots,k\}$.