We have a regular polygon. The triangle will be obtuse iff we connect one vertex with another vertex that is 4 or more vertices away, because then the opposite angle from that side would intercept an arc of greater than half the circle.
So this can be similarly stated as how many ordered partitions of $7$ intro three are there such that none of the partitions exceed $3$ and each is at least $1$. This is the same as how many solutions are there of $x_1 + x_2 + x_3 = 4$ with $x_i < 3$. There are ${6 \choose 4} = 15$ ways to do so; of these, $9$ of them do not fit the second restriction ($(3,1,0) \times 6; (4, 0, 0) \times 3$), giving us $6$.
Then we rotate each of these $6$ valid configurations $6$ times, giving $7 \times 6 = 42$ acute triangles. We triple count each, so there are $14$.
Alternate method:
There are $7 \choose 3$ triangles in total. If the three points lie on the same semi-circle, they will be obtuse. There are $7$ sets of $4$ consecutive points which lie on the same semi-circle, and for each set there are $3$ obtuse triangles we can make (from vertices $1-2-3, 1-2-4, 1-3-4$ -- note that $2-3-4$ is counted in the next set). Thus our answer is ${7 \choose 3} - 7(3) = 14$