I believe you can do something like this:
Let $X$ be zero mean. Then the covariance matrix is $  \text{cov}(X) = \mathbb{E}[XX^T]$. Hence, by the same argument:
\begin{align*}
  \text{cov}(X) = \text{arg} \min_{\Sigma} \; \mathbb{E} \| XX^T - \Sigma\|_F^2
\end{align*}
If $X$ not zero mean, let $X'$ be an independent copy. Then, $X - X'$ is zero mean an $\text{cov}(X-X') = 2 \text{cov}(X)$, hence
\begin{align*}
  \text{cov}(X) = \frac12\text{arg} \min_{\Sigma} \; \mathbb{E} \| (X-X')(X-X')^T - \Sigma\|_F^2.
\end{align*}