A regular 30-gon

Use the Pythagorean Theorem for this one. The diameter of the circle is $2$ , so that the sum of the squares of the non-hypotenuse sides of any right triangle with the diameter as the base is $4$ . Besides the diagonal that is the diameter itself, there are $14$ other diagonal pairs that make for the sides of such right triangles. Hence, the answer is $4(1+14) = 60$

Example, for diagonal $A_1A_{15}$ , choose diagonal $A_1A_2$ to make such a pair that will form a right triangle, with base $A_1A_{16}$ , etc.

Let $w=e^{2 \pi i/n}$ , where $n$ is the number of sides of the polygon. We want to find the sum: $\begin{aligned} S &= \sum_{k=0}^{n-1} |1-w^k|^2\\ &= \sum_{k=0}^{n-1} \left|1-\cos \dfrac{2 \pi k}{n}-i \sin \dfrac{2 \pi k}{n}\right|^2\\ &= \sum_{k=0}^{n-1} \left(\left(1-\cos \dfrac{2 \pi k}{n}\right)^2+\sin^2 \dfrac{2 \pi k}{n}\right)\\ &= \sum_{k=0}^{n-1} \left(2-2\cos \dfrac{2 \pi k}{n}\right)\\ &= 2n-2\sum_{k=0}^{n-1} \cos \dfrac{2 \pi k}{n}\\ &= 2n-2\text{Re}\left(\sum_{k=0}^{n-1} w^k\right) = 2n-2\text{Re}\left(\dfrac{w^n-1}{w-1}\right)\\ &= 2n-2\text{Re}\left(\dfrac{1-1}{w-1}\right) = 2n\\ \end{aligned}$ In this case we have $n=30$ , so $S=2(30)=\boxed{60}$ .

$Using ~Cos ~Law,~ we~ have~ 29~ diagonals.~~ So:-\\ \displaystyle \sum_{i=1}^{29}\left (1^2+1^2-2*1*1*Cos \left(\dfrac{2*\pi}{30}*i \right) \right )=60.$
I have used TI-83 Plus calculator with following pressed 29 times.
0 STO-> X ENTER
0 STO-> A ENTER
ENTER
ENTER
$1+X ~STO\rightarrow~X~ :~ A+2*\left (1 - Cos \left (\dfrac{X*2\pi}{30} \right ) \right )~ STO\rightarrow~A$ ENTER ENTER ENTER . . . ENTER....29 times.

Use simple trigonometry and get the sides of the polygon.Then add them up and use formula for trigonometric angles in Arithmetic progression.

Working up from equilateral triangle, square, regular pentagon, the pattern was compelling enough to chance a guess. No pen and paper available to attempt a proof!

What a pleasant and intriguing surprise! Thank you!