This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula
$$ F_n = (A^n-B^n)/(A-B), $$
where $A$ and $B$ are the roots of $T^2-T-1$. So if $\sqrt5$ is in your finite field, then so are $A$ and $B$, and since $AB=-1$, the period divides $p-1$ from Fermat's little theorem. If not, then you're in the subgroup of $\mathbb F_{p^2}$ consisting of elements of norm $1$, giving the other result. If you want small period, then take primes that divide $A^n-1$, or really its norm, so take primes dividing $(A^n-1)(B^n-1)$, where $A$ and $B$ are $\frac12(1\pm\sqrt5)$. An open question is in the other direction: Are there infinitely many $p\equiv\pm1\pmod5$ such that the period is maximal, i.e., equal to $p-1$?