A triple-sum Ramanujan-type supercongruence

Let $p>3$ be a prime number. Define the triple sum

$$S_p= \sum_{k=0}^{p-1} \sum_{j=0}^k \sum_{i=0}^{k-j} \frac{ {2j\choose j}^2{4j\choose 2j} {2i\choose i}^2{4i\choose 2i} {2(k-j-i)\choose k-j-i}^2{4(k-j-i)\choose 2(k-j-i)}}{2^{8k} 3^{2k}} (8i+1)(8j+1) \big(8(k-j-i)+1 \big).$$

Triple-sum Ramanujan-type congruence. One has

$$S_p\equiv 0 \pmod{p^2}.$$

This conjecture is motivated by the failure of the corresponding triple sum to satisfy the stronger congruence modulo $p^3$, despite the analogous double-sum congruence modulo $p^3$. It is based on computational evidence, and its general validity remains open.