Obviously, such turning is possible for $n = 1$. For $n = 2$, flip each of the four $1$-sided equilateral triangles once and all the coins will be tail-up.

We shall use now induction of step $3$. Assume that $n$ is an admissible value. Flipping the coins of each unit sided triangle of an equilateral triangle of side length $n+3$, the coins from the vertices of the big triangle will turn one time, those along the sides three times and the interior coins will turn six times each. Consequently, all the exterior coins are turned tail up and all the interior coins are heads up. But the interior coins form the net corresponding to an $n$-sided triangle, so the induction works.

If $3 \mid n$, then color the coins in red, yellow and blue so that any three adjacent coins have different colors. Also, any three coins in a row will have different colors. In this case the corners will all have the same color, say red. Since there are, in total, $\frac{(n+1)(n+2)}{2} \equiv 1 \pmod{3}$ coins, then there will be exactly one more red coin than yellow or blue ones. Thus, at the beginning, the parity of the number of red heads is different than the parity of the number of yellow heads. Since each move changes the parity of the number of heads of each color, we cannot end up with the parity of red heads equal to that of yellow or blue heads, which would be the case if all coins showed tails. Thus the coins cannot all be inverted, so $n$ is not an admissible value.