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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} \max_{\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} \max_{\Sigma} \; \mathbb{E} \| (X-X')(X-X')^T - \Sigma\|_F^2. \end{align*}