You are asking for $\text{Var}(\sum_i X_i)$ when $\sum_i X_i$ is a vector of multiple elements, though I think what you're asking for is the covariance matrix (the generalization of variance to a vector).
You can solve this in a similar way to the univariate case.
Suppose we have two variables, $x, y \in \mathbb{R}^1$, each with $n$ observations. We want to calculate
$$
\text{Cov}(\sum_ix_i, \sum_j y_j).
$$
Once you do that, then you can simply generalize to $x \in \mathbb{R}^d$ by generating the covariance matrix with each pairing of dimension using the same formula as above, for $x$ and $y$. So,
$$
\begin{split}
\text{Cov}(\sum_ix_i, \sum_j y_j) &= \left\langle \left( \sum_i x_i - \langle \sum_i x_i \rangle \right) \left( \sum_j y_j - \langle \sum_j y_j \rangle \right) \right\rangle \\
&= \sum_{i,j} \left\langle \langle x_i y_j \rangle + \langle x_i \rangle \langle y_j \rangle - x_i \langle y_j \rangle - \langle x_i \rangle y_j \right\rangle \\
&= \sum_{i,j} \left( \langle x_i y_j \rangle - \langle x_i \rangle \langle y_j \rangle\right) \\
&= \sum_{i,j} \text{Cov}(x_i,y_j),
\end{split}
$$
where the sum is over all $n^2$ pairings of $i,j$. In the second line, I used the fact that $\langle \sum_i x_i \rangle = \sum_i \langle x_i \rangle$, which is true even if the terms are not independent. To see this, express the expectation as an integral over the $n$-dimensional space of observations, with a joint distribution $p(x)$. Lack of independence means you can't split $p(x)$ into a product of individual distributions, but you can still pull the sum out of the integral, and the expectation of each term is simply taken with respect to the whole set of observations.
This is certainly symmetric because the indices that run over $x$ and $y$ are symmetric. It also reduces to the univariate case ($y=x$) because you can split the sum into the diagonal elements (the $\text{Var}$ terms) and twice the upper triangle elements (the $\text{Cov}$ terms for $i<j$, which is equivalent to the lower triangle terms). In the general case, however, you don't get to simply take the upper right triangle indices and double them, because $x_i y_j \ne x_j y_i$, so you have to sum over all pairings.
Thus, in general,
$$
\text{Cov}\left(\sum_i X_i^{(k)}, \sum_j X_j^{(m)} \right) = \sum_{i,j} \text{Cov} \left( X_i^{(k)}, X_j^{(m)} \right).
$$
Using the matrix computation only (although essentially, this solution is not that different from @DeltaIV's direct calculation), let me first slightly modify your definition of $S$ to its centralized version $\begin{bmatrix}\hat{\theta}_1 - \theta & \cdots & \hat{\theta}_m - \theta\end{bmatrix}$. We can go as follows
\begin{align}
& E[(\hat{\theta} - \theta)^T(\hat{\theta} - \theta)] \\
= & \frac{1}{m^2}E[1^TS^TS1] \\
= & \frac{1}{m^2}1^T E[S^TS] 1 \tag{1} \\
= & \frac{1}{m^2}1^T \begin{bmatrix} E[(\hat{\theta}_1 - \theta)^T(\hat{\theta}_1 - \theta)] & \cdots & E[(\hat{\theta}_1 - \theta)^T(\hat{\theta}_m - \theta)] \\
\vdots & \ddots & \vdots \\
E[(\hat{\theta}_m - \theta)^T(\hat{\theta}_1 - \theta)] & \cdots & E[(\hat{\theta}_m - \theta)^T(\hat{\theta}_m - \theta)] \end{bmatrix} 1 \\
= & \frac{1}{m^2}1^T\text{diag}(\sigma^2, \ldots, \sigma^2)1 \tag{2} \\
= & \frac{1}{m}\sigma^2.
\end{align}
In $(1)$, we used the fact that for any conformable non-random matrices $A$, $B$ and random matrix $X$, $E[AXB] = AE[X]B$.
In $(2)$, we applied the independence assumption.
Best Answer
It's useful to know how to convert the notation of Complex numbers into a more familiar Real notation. Here are the details.
Hermitian matrices and real-valued Complex forms
$\mathbf Q = (q_{ij})$ is a square matrix representing a real-valued quadratic form. This means that we don't really care about the details of $\mathbf Q:$ what matters is the function of the (Complex) vector $\mathbf z = (z_1, z_2, \ldots, z_n)^\prime$ defined via matrix multiplication by
$$Q(\mathbf z) = \mathbf z^{*}\, \mathbf Q\, \mathbf z = \sum_{i=1}^n \sum_{j=1}^n \bar z_i\, q_{ij}\, z_j.\tag{1}$$
Writing $z_i = x_i + y_i\mathbf i$ in terms of its real and imaginary parts, the overline refers to the complex conjugate $\bar z = x_i - y_i\mathbf i$ and the conjugate transpose operation transposes a vector and conjugates all its components,
$$\mathbf z^* = (\bar z_1, \bar z_2, \ldots, \bar z_n).$$
Because $Q$ is required to be a real number for all such vectors $\mathbf z,$ it must equal its conjugate, whence (by swapping the symbols "$i$" and "$j$" in the sums) we see
$$Q(\mathbf z) = \overline {Q(\mathbf z)} = \sum_{i=1}^n \sum_{j=1}^n z_i\, \bar q_{ij}\, \bar z_j = \sum_{j=1}^n \sum_{i=1}^n z_j\, \bar q_{ji}\, \bar z_i$$
Comparing this to $(1)$ shows that for equality to hold for all $\mathbf z,$ necessarily $q_{ij} = \bar q_{ji}$ for all indexes $i$ and $j.$ Such a matrix is called Hermitian and this operation of transposing a matrix while taking complex conjugates is evidently a generalization of the conjugate transpose, allowing us to write
$$\mathbf Q^{*} = (\bar q_{ji}) = \overline{\mathbf Q}^{\prime}.$$
In other words,
Writing Complex forms as equivalent Real forms
Write $q_{ij} = a_{ij} + b_{ij}\mathbf i.$ The Hermitian property of $\mathbf Q$ is
$$a_{ji} - b_{ji}\mathbf i = \bar q_{ji} = q_{ij} = a_{ij} + b_{ij}\mathbf i.$$
Comparing real and imaginary parts and assembling them into matrices $\mathbf A = (a_{ij})$ and $\mathbf B = (b_{ij})$ shows
$$\mathbf A = \mathbf A^\prime\ \text{and}\ \mathbf B = -\mathbf{B}^\prime.\tag{2}$$
$\mathbf A = \Re (\mathbf Q)$ is the real part of $\mathbf Q$ and $\mathbf B = \Im (\mathbf Q)$ is its imaginary part. If these (simple) relationships are not perfectly clear, see the concrete example (with $n=2$) offered at the end of this post.
Let $i$ and $j$ be arbitrary indexes. Let's examine how the real and imaginary parts of $z_i$ and $z_j$ appear in $Q(\mathbf z).$ This product appears in at most two terms in the sum $(1),$ which can be combined using the Hermitian property of $\mathbf Q.$ Here is the calculation when $i\ne j.$ We only have to compute the real part of the result, knowing the imaginary part will be zero.
$$\begin{aligned} \bar z_iq_{ij}z_j + \bar z_jq_{ji}z_i &= (x_i - y_i\mathbf i)(a_{ij} + b_{ij}\mathbf i)(x_j + y_j\mathbf i) + (x_j - y_j\mathbf i)(a_{ji} + b_{ji}\mathbf i)(x_i + y_i\mathbf i)\\ &= (x_i - y_i\mathbf i)(a_{ij} + b_{ij}\mathbf i)(x_j + y_j\mathbf i) + (x_j - y_j\mathbf i)(a_{ij} - b_{ij}\mathbf i)(x_i + y_i\mathbf i)\\ &= 2x_i\, a_{ij}\, x_j + 2y_i\, a_{ij}\, y_j - 2x_i\, b_{ij}\, y_j + 2 x_j\, b_{ij}\, y_i \end{aligned}$$
(I leave the calculation for $i=j$ as a simple exercise.)
This is a quadratic form in the $2n$ dimensional real vector $\mathbf z_{\Re} = (x_1,x_2,\ldots, x_n;y_1,y_2,\ldots,y_n).$ In block matrix notation its matrix is
$$\mathbf Q_{\Re} = \pmatrix{\mathbf A & -\mathbf B \\ \mathbf B & \mathbf A}.\tag{3}$$
Because $\mathbf A$ is symmetric and $\mathbf B$ is antisymmetric (according to $(2)$ above), evidently $\mathbf Q_{\Re}$ is a real, symmetric matrix -- as it must be, because we have already seen it represents a real-valued quadratic form in the real variables $x_1, \ldots, y_n.$
Solution to the problem in the question
Our thread at Variance of quadratic form for multivariate normal distribution shows that
$$\operatorname{Var}\left(\mathbf z^* \mathbf Q\, \mathbf z\right) = \operatorname{Var}\left(\mathbf z_{\Re}^\prime\, \mathbf Q_{\Re}\, \mathbf z_{\Re}\right) = 2\operatorname{tr}\left(\mathbf Q_{\Re}^2\right).$$
The block matrix expression for this form's matrix enables us to compute this trace in terms of the real and imaginary parts of $\mathbf Q,$
$$\operatorname{tr}\left(\mathbf Q_{\Re}^2\right) = \operatorname{tr}\left(\pmatrix{\mathbf A & -\mathbf B \\ \mathbf B & \mathbf A}^2\right) = \operatorname{tr}\pmatrix{\mathbf A^2 - \mathbf B^2 & -2\mathbf A \mathbf B\\2\mathbf B \mathbf A & \mathbf A^2 - \mathbf B^2} = 2\operatorname{tr}\left(\mathbf A^2\right) - 2\operatorname{tr}\left(\mathbf B^2\right).$$
Thus,
Checking the result
I generated a random $2\times 2$ Complex covariance matrix $\mathbf Q,$ computed $Q(\mathbf z)$ for $100$ iid standard Normal complex vectors $\mathbf z,$ and estimated their variance. Upon repeating this five thousand times I obtained an empirical representation of this distribution. If the preceding result is correct, then the ratio of each of these variance estimates to the calculated variance should be centered at zero. Here is the histogram
It's exactly as expected. I repeated this process several times for $2\times 2$ through $5\times 5$ Complex covariance matrices and observed comparable results.
To illustrate the block matrix notation used above, here are the computer representations of $\mathbf Q$
and $\mathbf Q_{\Re}$
for the original simulation. Comparing them will make their relationship clear.
Here is the
Rcode used for this check.