Let's define the nonnegative random variable $X$ as $$ X := |\hat \theta - \theta| \geq 0 .$$

Markov's inequality states that: if $X$ is a nonnegative random variable and $a > 0$, then $$ \operatorname {Pr} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}. $$

Therefore:

Result 1 $$ {\operatorname {E}[ |\hat \theta - \theta| ]} \geq a \cdot \operatorname {Pr} (|\hat \theta - \theta| \geq a). $$

A second result uses the corollary to Markov on the quantile function which states that: for a nonnegative random variable $X$, the quantile function of $X$ satisfies $$ Q_{X}(1-p) \leq {\frac {\operatorname {E} (X)}{p}}, \quad 0 < p < 1. $$ Therefore:

Result 2 $$ {\operatorname {E} [|\hat \theta - \theta|]} \geq p \cdot Q_{|\hat \theta - \theta|}(1-p), \quad 0 < p < 1. $$

Let's define the nonnegative random variable $X$ as $$ X := |\hat \theta - \theta| \geq 0 .$$

Markov's inequality states that: if $X$ is a nonnegative random variable and $a > 0$, then $$ \operatorname {Pr} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}. $$

Therefore:

Result 1 $$ {\operatorname {E}[ |\hat \theta - \theta| ]} \geq a \cdot \operatorname {Pr} (|\hat \theta - \theta| \geq a). $$

A second result uses the corollary to Markov on the quantile function which states that: for a nonnegative random variable $X$, the quantile function of $X$ satisfies $$ Q_{X}(1-p) \leq {\frac {\operatorname {E} (X)}{p}}, \quad 0 < p < 1. $$ Therefore:

Result 2 $$ {\operatorname {E} [|\hat \theta - \theta|]} \geq p \cdot Q_{|\hat \theta - \theta|}(1-p), \quad 0 < p < 1. $$

Note: Regretfully, I do not currently have an idea how to derive any asymptotic formula, neither $\Theta(1/n)$ nor $\Theta(1/ \sqrt n)$, for any of my 2 results. All suggestions are welcome.