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][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 $$ \left(\{\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}}  \sum_{i=1}^{2^n} Z_{n,i,j}\neq 0 \}\right)\subset\bigcup_{i,j=1}^{2^n}\left(\{Y_{i,j}>\varepsilon 2^{2n}\}\right).$$
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[1] Deviation inequalities for Banach space valued martingales differences sequences and random fields Davide  Giraudo ESAIM: PS 23 922-946 (2019)

 


  [1]: https://www.esaim-ps.org/articles/ps/pdf/2019/01/ps180007.pdf