Let $\{X_1, X_2, \ldots, X_n\}$ be independent and identically distributed (i.i.d.) random variables sampled from a common distribution with density $f_{\theta}(x)$, where $\theta$ is an unknown parameter. We want to estimate $\theta$ given these $n$ samples. Suppose $\hat{\theta}$ is an estimator based on these samples. For simplicity, suppose this is unbiased, so that $E[\hat{\theta}] = \theta$.

Cramer-Rao bound theory implies that for any unbiased estimator:
$$ E[(\hat{\theta} - \theta)^2] \geq \frac{1}{I(\theta)} = \Theta(1/n) $$
where $I(\theta)$ is the Fisher information.
However, I am interested not in the mean-square error, but the *mean absolute error*:
$$ E[|\hat{\theta} - \theta|] \geq ??? $$

This must be a well-studied problem. Any references or insights on this would be helpful.

Intuitively one expects $E[|\hat{\theta}-\theta|]\geq \Theta(1/\sqrt{n})$, and this is what I eventually want to show for my particular context (actually, eventually I am interested in possibly biased estimators). If one assumes the absolute error is at most $M$ then: $$ \Theta(1/n) \leq E[(\hat{\theta}-\theta)^2] \leq ME[|\hat{\theta}-\theta|] $$ but this inequality is weaker than I want since it means the absolute error also decays by at most $\Theta(1/n)$, whereas I want to increase the bound to $\Theta(1/\sqrt{n})$.

Actually, I can prove something of this form in a special case when $\theta$ represents the mean $E[X_1]$. I'm wondering if such a thing is known? Estimating the mean leads to the "obvious" estimator $\hat{\theta}=\frac{1}{n}\sum_{i=1}^nX_i$, but it is not obvious how to show this is "best" in some sense, particularly for the mean-absolute-error.