The minimzer of the MSE $E[(a-X)^2]$ is $a=E(X)$, and the MSE can be decomposed into $E[(a-X)^2] = E[(a-E(X))^2] + Var(X)$.
I am wondering whether there exists a similar expression th MAE $E[|a-X|]$ in terms of its own minimizer $med(X)$?
Is there a known standard relationship between $E[|a-X|]$, $E[|a-med(X)|]$, and perhaps something like $E[|med(X)-X|]$? Or between $med[|a-X|]$, $med[|a-med(X)|]$, and $med[|med(X)-X|]$? Or anything else that resembles $E[(a-X)^2] = E[(a-E(X))^2] + Var(X)$?
The short answer to your question is: an analogous decomposition does exist and can be used to show that the minimizer of $\Delta(a) := E[|X - a|]$ is the median (see remarks below for the latter claim).
Denote the median of $X$ by $m$. Using the definition of expectation, we have (note that the essence of the proof, which is shared by the $L^2$ expectation decomposition, is "subtract-then-add"): \begin{align} & E[|X - a|] = \int_{-\infty}^a (a - x)dF(x) + \int_a^{\infty}(x - a)dF(x) \\ =& \int_{-\infty}^m(a - x)dF(x) + \int_m^a(a - x)dF(x) + \int_a^m(x - a)dF(x) + \int_m^\infty(x - a)dF(x) \\ =& \int_{-\infty}^m(m - x)dF(x) + \int_{-\infty}^m(a - m)dF(x) \\ &+ 2\int_a^m(x - a)dF(x) \\ &+ \int_m^\infty(x - m)dF(x) + \int_m^\infty(m - a)dF(x) \\ =& E[|X - m|] + (m - a)(P[X > m] - P[X \leq m]) + 2\int_a^m(x - a)dF(x) \\ =& E[|X - m|] + 2\int_a^m(x - a)dF(x). \end{align} In the penultimate step, we used the condition $P[X \leq m] = P[X > m] = 0.5$. Therefore, the decomposition is \begin{align} E[|X - a|] = E[|X - m|] + 2\int_a^m(x - a)dF(x). \tag{1} \end{align}
Note that the second term in the right hand side of $(1)$, which resembles the term $(E[X] - a)^2$ term in the $L^2$ decomposition, is always non-negative. Under the assumption that $F(x)$ is strictly increasing (so that $m$ is uniquely determined), it is immediate that $\Delta(a)$ is minimized at $a = m$ (for otherwise the integral is strictly positive). When the theoretical median is not unique, the minimizer of $\Delta(a)$ is not unique either.
For a general quantile position $\tau \in (0, 1)$ and the check function $\rho_\tau(u) := u(\tau - I_{(-\infty, 0)}(u))$, the similar decomposition to $(1)$ also holds as follows with the median $m$ replaced with the $\tau$-quantile $q_\tau$ (the proof is almost identical as above):
\begin{align} E[\rho_\tau(X - a)] = E[\rho_\tau(X - q_\tau)] + \int_a^{q_\tau}(x - a)dF(x). \tag{2} \end{align}
Note that $(1)$ and $(2)$ differ by a scaling factor $2$ because $|u| = 2\rho_{0.5}(u)$. Using $(2)$, it is also easy to conclude that $q_\tau = \operatorname{argmin}_{a \in \mathbb{R}^1}E[\rho_\tau(X - a)]$ given the same monotonicity condition of $F$.
