A Farey sequence of order $n$ is a list of the rational numbers between 0 and 1 inclusive whose denominator is less than or equal to $n$.
For example $F_6= \{0,1/6,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,5/6,1\}$.
The consecutive differences of $F_6$ are $S_6=\{1/6, 1/30, 1/20, 1/12, 1/15, 1/10, 1/10, 1/15, 1/12, 1/20, 1/30, 1/6\}$,
and the sum of squares of the elements of $S_6$, $\displaystyle T_6=\sum_{i\in S_6}i^2 = \frac{19}{180}$.
It appears that the sum of squares $T_n$ shrinks a bit faster than $O(ln(n)/n^2)$ but I cannot see how to prove it. I'd also be interested in considering higher powers than the square, and the differences between every other element, or every third, etc.
Actually, we can do it for higher powers as well. Let $F_k (N)$ denote the sum of the $k^{th}$ powers of the differences in the Farey sequence. Then it is easy to see that $$F_0(N)=\sum_{n=1}^N \phi(n)\sim \frac{3N^2}{\pi^2}$$ and $$F_1(N)=1.$$ It seems you are looking for $F_2 (N)$, and, you are correct that the asymptotic is $\frac{\log N}{N^2}$. More precisely we have: $$F_{2}(N)=\frac{12}{\pi^{2}N^{2}}\left(\log N+\frac{1}{2}+\gamma-\frac{\zeta'(2)}{\zeta(2)}\right)+O\left(\frac{\log^{2}N}{N^{3}}\right).$$
For $k\geq 3$, we the log disappears, and we have $$F_{k}(N)=\frac{2\zeta(k-1)}{\zeta(k)N^{k}}+O\left(\frac{\log N}{N^{k+1}}\right).$$ (The error can be improved by a log when $k\geq 4$)
Hope that helps,
References:
R.R. Hall's Paper: A Note on Farey Sequences.
Remark: I strongly suggest reading R.R. Hall's paper since it has the complete proofs, and covers other cases such as negative powers.