$\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\Cov}{\mathrm{Cov}}$ Extending Kevin's answer let me show that
$$S_{k,n} \leq \frac{k^3-k}{12}.$$
It is easy to see that $$\Var(X) \leq \E\[(X-c)^2\]$$ for any random variable $X$ and any constant $c$.
Let $a_\{i,j\}$ denote the entry of the matrix on the intersection of $i^\mathrm{th}$ row and $j^\mathrm{th}$ column. Let $R_i$ (resp. $C_j$) be the random variables corresponding to uniformly sampling entries of $i^\mathrm{th}$ row (resp. $j^\mathrm{th}$ column). Then
\begin{align*} S_{k,n} &= \max \sum_{i=1}^k \Var(R_i) \leq \sum_{i=1}^k \E\[(R_i - \frac{k+1}{2})^2\]=\frac{1}{n}\sum_{1\leq i \leq k, 1 \leq j \leq n} ( a_{ij}-(k+1)/2)^2 \newline &= \frac{k}{n}\sum_{j=1}^n \Var(C_j)= \frac{k}{n} \cdot n \cdot \frac{k^2-1}{12} \end{align*}
It is clear that the bound is achieved if and only if the average value of each row is $(k+1)/2$. Examples with this property can be constructed for all even $n$. It is also easy to see that for fixed $k$ and $n \to \infty$ one can make the row averages be arbitrarily close to $(k+1)/2$ and thus the total variance arbitrarily close to the upper bound.
If $k$ is even and $n$ is odd then the bound can not be achieved exactly, as $n(k+1)/2$ is not integral. When $k$ is odd and $n=3$ then one can have $i^\mathrm{th}$ row consisting of $$\langle i; (i + (k-1)/2) \:\mathrm{mod}\: k; (k-2i +2) \:\mathrm{mod}\: k \rangle.$$ Combining this construction with pairs of ``reverse" columns, one achieves the bound for all $n \geq 3$. To summarize:
- $S_{k,n} = \frac{k^3-k}{12}$, when $n$ is even, or $k$ is odd and $n \geq 3$;
$\max S_{k,n} = \frac{k^3-k}{12}$, when $n$ is even, or $k$ is odd and $n \geq 3$;
- $S_{k,n} < \frac{k^3-k}{12}$, otherwise, but $\lim_{n \to \infty} S_{k,n}=\frac{k^3-k}{12}$ for all $k$.
$\max S_{k,n} < \frac{k^3-k}{12}$, otherwise, but $\lim_{n \to \infty} \max S_{k,n}=\frac{k^3-k}{12}$ for all $k$.