Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.
However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.
That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ which is $\binom{n+2}{3} + \binom{n-1}{3}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.