There's no such proof, because we can get $\binom{n-2}{3}$ edges with no size-one intersection: just take the $5$-uniform hypergraph of all edges containing two specific vertices.
What about the upper bound? Well, a generalization of the Erdős–Ko–Rado theorem says that this is optimal for a slightly stronger condition: if $n$ is sufficiently large, there can be at most $\binom{n-2}{3}$ edges in a hypergraph where any two edges intersect at least twice. (In our case, two edges can also be disjoint.)
I suspect that $\binom{n-2}{3}$ is the best possible in this problem, as well, but cannot prove it. Here is a proof of an $O(n^3)$ upper bound which almost certainly overcomplicates things and gives away too much.
Suppose that $M = \{e_1, e_2, \dots, e_k\}$ is a largest family of pairwise disjoint edges in the hypergraph. The rest must intersect at least one of these. There are at most $\binom 52^2 \binom k2 n \le 100 \binom{n/5}{2} n < \frac12 n^3$ edges that intersect two edges in $M$, because then only one vertex of those edges is left to pick freely. Delete these, and label each edge (including the edges in $M$) with a number $1, \dots, k$ according to which edge in $M$ it intersects.
Let $x$ be a vertex outside $\bigcup M$ contained in some edge $f$ with label $i$. Then any edge with a label $j \ne i$ containing $x$ must contain a second vertex of $f$ outside $e_i$, two vertices of $e_j$, and one other vertex: there are at most $20n$ choices total. So if we say that a vertex $x$ also has label $i$ if it's contained in more than $20n$ edges with label $i$, then a vertex can have at most one label. To clean up, for each vertex $x$ outside $\bigcup M$, delete the edges containing $x$ whose label is not the label of $x$, and for every vertex $x$ with no label, delete all edges containing it; this removes at most $(n-5k)k(20n) \le 4n^3$ edges total.
Finally, partition all vertices into $S_1, S_2, \dots, S_n$ according to their label, putting the vertices of $e_i$ into $S_i$ as well. The edges of label $i$ form a $2$-intersecting family contained in $S_i$: two edges of label $i$ cannot be disjoint, because then we could get a larger disjoint family than $M$ by replacing $e_i$ with those two edges. Therefore, by the Erdős–Ko–Rado extension cited above, at most $\binom{|S_i|-2}{3}$ edges have label $i$. By convexity, $$\binom{|S_1|-2}{3} + \dots + \binom{|S_k|-2}{3}$$ is maximized when all of the $S_i$ are empty except for one, in which case it is equal to $\binom{n-2}{3}$.
Taking into account our deletions, our original hypergraph had at most $\binom{n-2}{3} + 4n^3 + \frac12 n^3$ edges, which is $O(n^3)$.
