We have the numbers $2,3,4,5,6,7$. Recall that the sum of any two sides of a triangle must be greater than the third side. Equivalently, the sum of the two shortest sides must be greater than the third side.
We are making a triangle. The shortest stick we use can be any of $2,3,4,5$. We find for each choice of shortest stick, how many bad choices there are for the other two sticks. (Bad choice = No triangle.)
Shortest stick is 2: If next shortest stick is $3$, then $5$, $6$, and $7$ for third are bad.
If next shortest after $2$ is $4$, then $5$ and $6$ for third are bad.
If next shortest after $2$ is $5$, then $7$ for third is bad.
Total bad here: $3+2+1=6$.
Shortest stick is 3: You can count the number of bads of this type.
Shortest is 4 or 5: No bads.
Now conclusion can be reached.
There are at least two different ways to extend this problem to larger $n$. Here's an approach which works for the easier of the two ways, which is where the vertices of $K_n$ are $n$ points in general position forming a convex $n$-gon, so every internal node is on exactly two edges.
Every triangle can be associated with a triple of distinct edges of $K_n$ (the edges you get if you extend the sides of the triangle as far as possible), and each triple of edges is associated with at most one triangle. So we just need to count the "good" triples which do give a triangle, i.e. every pair of edges meets (either at a vertex of the original graph or at an internal node).
Suppose we know the set of endpoints of a good triple of edges (without multiplicity): how many good triples have exactly those endpoints? If there are six points in the set, $a,b,c,d,e,f$ in cyclic order, there is only one good triple consisting of the edges $ad,be,cf$. If we have five endpoints, $a,b,c,d,e$ in cyclic order, there is exactly one point in two of the edges. Suppose it is $a$. Then the only possibility is $ac,ad,be$ (thanks to Henning Makholm for pointing out that my other "possibilities" weren't valid). Since the choice of $a$ was arbitrary, there are $5$ possibilities in total. If there are four points $a,b,c,d$ in the set, the only possibilities are $ab,bd,ac$ and cyclic rearrangements. Finally, if there are only three points there is only one possibility.
This means the number of triangles is $\binom n6+5\binom n5+4\binom n4+\binom n3$.
To deal with the other plausible extension (the vertices form a regular $n$-gon) you'd need to calculate the number of degenerate triangles arising from three diagonals which all meet at a point, and subtract them off.
Best Answer
I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.
You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $\lfloor28/3\rfloor=9$.
I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work: $$ABC,\ ADG,\ AFH,\ BEH,\ BFG,\ CDH,\ CEG,\ DEH$$
P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $n\le3p$ (since at most $3$ triangles can contain a given point), so $t\le p=8$.
P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3\times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.