A set of such polyhedra can be obtained from a dodecadron through repeated edge truncation: in Conway's notation, that is the chamfer transform denoted through c. Have a look at this wonderful site for generating pictures of such objects:
Obviously three hexagons sharing a common vertex cannot be perfectly regular, otherwise the surface would be flat in a neighbourhood of the common vertex!
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
The issue is that the picture depicts not the conventional soccer ball (a truncated icosahedron) but rather something a little different, the chamfered dodecahedron, also known as a truncated rhombic triacontahedron. This actually does have 120 edges.
So, in a sense, you were right both times, but just thinking about different polyhedra!
It's interesting to note that these are both examples of Goldberg Polyhedra, polyhedra made from only pentagons and hexagons -- although the faces are not necessarily regular (and in the chamfered dodecahedron, they are not).