Let the $i$-th column and the $j$-th row have $a_i, b_j$ white cells respectively, and let $R$ be the set of red cells. For a red cell with coordinates $(i, j)$, there are $a_i b_j$ feasible sequences $(C_1, C_2, C_3)$ with $C_2 = (i, j)$. Therefore
$$T = \sum_{(i,j) \in R} a_i b_j.$$
Using the inequality $2ab \le a^2 + b^2$, we obtain
$$T \le \frac{1}{2} \sum_{(i,j) \in R} (a_i^2 + b_j^2) = \frac{1}{2} \sum_{i=1}^{n} (n-a_i)a_i^2 + \frac{1}{2} \sum_{j=1}^{n} (n-b_j)b_j^2, \quad (1)$$
this is because in the $i$-th column there are $n-a_i$ red cells, and in the $j$-th row there are $n-b_j$ red cells. Now we find the maximum value of expression (1).
By the AM-GM inequality, we have
$$(n-x)x^2 = \frac{1}{2}(2n-2x) \cdot x \cdot x \le \frac{1}{2}\left(\frac{2n}{3}\right)^3 = \frac{4n^3}{27},$$
where equality holds if and only if $x = \frac{2n}{3}$. Combining all the expressions, we get
$$T \le \frac{n}{2} \cdot \frac{4n^3}{27} + \frac{n}{2} \cdot \frac{4n^3}{27} = \frac{4n^4}{27}.$$
If $n = 999$, then any coloring satisfying that each column and each row has $x = \frac{2n}{3} = 666$ white cells will achieve the above maximum value of $T$, because all the inequalities above become equalities. For example:

Color cell $(i, j)$ white if $i - j \equiv 1, 2, \dots, 666 \pmod{999}$, and red otherwise; this is a feasible coloring.

In summary, the maximum value that $T$ can attain is $\frac{4 \cdot 999^4}{27}$.