We prove that in an $n \times n$ square table there are at most $\frac{4 n^{4}}{27}$ such triples.
Let row $i$ and column $j$ contain $a_{i}$ and $b_{j}$ white cells respectively, and let $R$ be the set of red cells. For every red cell $(i, j)$ there are $a_{i} b_{j}$ admissible triples $(C_{1}, C_{2}, C_{3})$ with $C_{2}=(i, j)$, therefore
$$T=\sum_{(i, j) \in R} a_{i} b_{j}$$
We use the inequality $2 a b \leq a^{2}+b^{2}$ to obtain
$$T \leq \frac{1}{2} \sum_{(i, j) \in R}\left(a_{i}^{2}+b_{j}^{2}\right)=\frac{1}{2} \sum_{i=1}^{n}\left(n-a_{i}\right) a_{i}^{2}+\frac{1}{2} \sum_{j=1}^{n}\left(n-b_{j}\right) b_{j}^{2} .$$
This is because there are $n-a_{i}$ red cells in row $i$ and $n-b_{j}$ red cells in column $j$. Now we maximize the right-hand side.
By the AM-GM inequality we have
$$(n-x) x^{2}=\frac{1}{2}(2 n-2 x) \cdot x \cdot x \leq \frac{1}{2}\left(\frac{2 n}{3}\right)^{3}=\frac{4 n^{3}}{27}$$
with equality if and only if $x=\frac{2 n}{3}$. By putting everything together, we get
$$T \leq \frac{n}{2} \frac{4 n^{3}}{27}+\frac{n}{2} \frac{4 n^{3}}{27}=\frac{4 n^{4}}{27}$$
If $n=999$ then any coloring of the square table with $x=\frac{2 n}{3}=666$ white cells in each row and column attains the maximum as all inequalities in the previous argument become equalities. For example color a cell $(i, j)$ white if $i-j \equiv 1,2, \ldots, 666(\bmod 999)$, and red otherwise.
Therefore the maximum value $T$ can attain is $T=\frac{4 \cdot 999^{4}}{27}$.