@FrancescoPolizzi could you please elaborate on how "blow-ups" distinguish, say, $t^5$ and $t^7$ near $t=0$?

By manipulating the "generalized cossine law" I put, it's possible to prove that if the segment on the extremity has a greater length than its adjacent segment, then swapping them will always increase the value of the distance. Maybe this can help us prove that at the endpoints of the segment list we have either a 0 or a 1

For the angles, if we consider the simmetry, permutations do seem unimodal. For $n \geq 4$, we seem to have $2^{n-4}+1$ permutations. Of these, $2^{n-4}$ are unimodal ones with endpoints $n-2$ and $n-3$. The other one has endpoints $n-2$ and $n-4$

I updated the link. Now the arrangements are better. Also, I noticed that there are $2^{n-3}$ possible arrangements of the $l$'s

Sorry, I messed the order of the angles. You're supposed to rotate by $\alpha_1$ first (i.e. put $l_0$ rotate $\alpha_1$, put $l_2$ rotate $\alpha_0$ and then put $l_1$). With this, you go from (0, 0) to (1, 0). Then to (3.4, 1.8). Then to (4.6, 3.4). So the distance squared is 32.72 > 32.32.

@MattF. Actually, this case ($n=3$) has already been completely solved: as stated in my question, no matter what are the values of the fixed and given segment lengths and angles, we have that, out of the $\dfrac{3! 2!}{2}$ possible arrangements, the following is always the one which maximizes the distance. If $l_0 < l_1 < l_2$ are the lengths and $\alpha_0 < \alpha_1$ the angles, the best arrangement is always built by putting $l_0$ first, rotating by $\alpha_0$, then putting $l_2$, rotating by $\alpha_1$ and then putting $l_1$. So for $n=3$ his algorithm always fails

It also fails for the solved n=3 case

@JosephO'Rourke surprisingly, this already fails for n=4. What I observed was a central tendency in which longer lengths and small angles tend to be in the middle. For n=4, the 3 possible cases I found were 1) ls 0 2 3 1 angs 2 0 1 || 2) ls 0 3 2 1 angs 2 0 1 || 3) ls 0 2 3 1 angs 1 0 2 || (for example (1) means that l_0 (the lowest) is put first, then rotate by alpha_2 (the biggest angle), put l_2, rotate by alpha_0 (the smallest angle), put l_3 (the biggest length), rotate by alpha_1 (the middle angle) and then put l_1 (the second smallest length) )