Let $A_{i, j}$ be the entry in the $i^{\text {th }}$ row and the $j^{\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that
$$\sum_{i=1}^{n} R_{i} \equiv(n-1)+P \quad\left(\bmod n^{4}\right)$$
Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion.
By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence
$$R_{i}=\prod_{j=1}^{n}\left(1+a_{i, j}\right)=1+\sum_{j=1}^{n} a_{i, j}+\sum_{1 \leqslant j_{1}<j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}}+\cdots \equiv 1+\sum_{j=1}^{n} a_{i, j} \equiv 1-n+\sum_{j=1}^{n} A_{i, j} \quad\left(\bmod n^{2}\right)$$
for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\left(\bmod n^{2}\right)$, and so $n^{2} \mid r_{i}$.
Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain
$$P=\prod_{i=1}^{n} R_{i}=\prod_{i=1}^{n}\left(1+r_{i}\right) \equiv 1+\sum_{i=1}^{n} r_{i} \quad\left(\bmod n^{4}\right)$$
whence
$$\sum_{i=1}^{n} R_{i}=n+\sum_{i=1}^{n} r_{i} \equiv n+(P-1) \quad\left(\bmod n^{4}\right)$$
as desired.
Comment. The original version of the problem statement contained also the condition
(iii) The product of all the numbers in the table is congruent to 1 modulo $n^{4}$.
This condition appears to be superfluous, so it was omitted.