Chords have length $2\sin(k\pi/9)$ for $k=1,\dots,8$, with $k$ and $9-k$ giving equal lengths, so there are $4$ distinct lengths each occurring $9$ times.
As in the problem, from one vertex the sum of fourth powers is $\sum_{k=1}^8\left(2-2\cos\frac{2k\pi}9\right)^2=\sum_{k=1}^8\left(4-8\cos\frac{2k\pi}9+4\cos^2\frac{2k\pi}9\right)=32+8+4\cdot\frac72=54$, using $\sum_{k=1}^8\cos\frac{2k\pi}9=-1$ and $\sum_{k=1}^8\cos^2\frac{2k\pi}9=4-\frac12$.
Each segment is counted from both ends, so the total is $\frac{9\cdot54}2=\boxed{243}$.
