In the Math Stack Exchange post, I gave a proof based on Lemma 2 in Bai and Yin (1993). I will give an alternative proof.
Expressing $\sum_{1\leqslant i\neq i'\leqslant n}X_{i,j}X_{i',j}$ as $\left(\sum_{i=1}^n X_{i,j}\right)^2-\sum_{i=1}^nX_{i,j}^2$ and considering dyadic numbers, we are reduced to show that $$ \tag{1}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,j}\right\rvert\to 0\mbox{ almost surely} $$ $$ \tag{2}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} X_{i,j}^2\to 0\mbox{ almost surely}. $$ By the Borel-Cantelli lemma, in order to prove (1), it suffices to show that $$ \tag{1'}\forall\varepsilon>0, \sum_{n\geqslant 1}\mathbb P\left(\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,j}\right\rvert>\varepsilon\right) <\infty$$
which reduces, by a union bound, to prove that $$ \tag{1''}\forall\varepsilon>0, \sum_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,0}\right\rvert>\varepsilon\right) <\infty$$
Let us prove (1''). Using Corollary 1.5 in 1 (applied with $r'=r=2$ and $B=\mathbb R$, hence $C_{r',B}=1$, we know that for each $q>0$, there exist constants $A(q)$ and $B(q)$ such that for each $x>0$ and each independent sequence $(Y_i)$, $$ \mathbb P\left(\max_{1\leqslant \ell\leqslant N}\left\lvert \sum_{i=1}^\ell Y_i\right\rvert >x\right)\leqslant A(q)\int_0^1u^{q-1}\mathbb P\left(\max_{1\leqslant i\leqslant N}\lvert Y_i\rvert>xB(q)u\right)du+A(q)x^{-q}\left(\mathbb E\left[Y_i^2\right]\right)^{q/2}. $$
Applying this inequality to $Y_i=X_{i,0}$, $x=2^n\varepsilon$ and $q=3$ shows that $$ \sum_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,0}\right\rvert>\varepsilon\right)\leqslant A(3)\sum_{n\geqslant 1}2^{2n}\int_0^1u^{2}\mathbb P\left( \lvert X_{i,0}\rvert>\varepsilon B(3)2^nu\right)du + A(3)\sum_{n\geqslant 1}2^{n}2^{-3/2}\varepsilon^{-3/2} $$ and finiteness of the first series follows from the elementary fact that $\sum_{n\geqslant 1}2^{2n}\mathbb P\left(Y>2^n\right)\leqslant 4\mathbb E\left[Y^2\right]$.
(2'') follows from an application of the similar argument as the usual strong law of large number, but the truncation level changes. Let $Y_{i,j}=X_{i,j}^2$ and for a fixed $\varepsilon$ and $n$, let $$ Z_{n,i,j}=Y_{i,j}\mathbf{1} \{Y_{i,j}\leqslant\varepsilon 2^{2n} \}-\mathbb E\left[Y_{i,j}\mathbf{1}\{Y_{i,j}\leqslant\varepsilon 2^{2n}\}\right], $$ $$ W_{n,i,j}=Y_{i,j}\mathbf{1}\{Y_{i,j}>\varepsilon 2^{2n}\}-\mathbb E\left[Y_{i,j}\mathbf{1}\{Y_{i,j}>\varepsilon 2^{2n}\}\right], $$ We thus have to show that $$ \tag{2'}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\to 0\mbox{ almost surely}. $$ $$ \tag{2''}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\to 0\mbox{ almost surely}. $$ For (2'), it suffices to show (by the Borel-Cantelli lemma) that $\sum_m\mathbb E\left[\left(\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\right)^2\right]<\infty$. For (2''), note that $$ \left(\{\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\neq 0 \}\right)\subset\bigcup_{i,j=1}^{2^n}\left(\{Y_{i,j}>\varepsilon 2^{2n}\}\right).$$
1 Deviation inequalities for Banach space valued martingales differences sequences and random fields Davide Giraudo ESAIM: PS 23 922-946 (2019)
2 L. Rüschendorf, Ordering of distributions and rearrangement of functions. Ann. Probab. 9