Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.
Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has the discrete distribution
$Z = -0.5$, with probability (wp) $.25$
$ = +0.5$, wp $.25$
$ = -1.2$, wp $.25 - \theta/2$
$ = +1.2$, wp $.25 - \theta/2$
$ = -\sqrt{0.155/\theta + 1.44}$, wp $\theta/2$
$ = +\sqrt{0.155/\theta + 1.44}$, wp $\theta/2$.
The family of distributions of $X$ is indexed by three parameters: $\mu$, $\sigma$, and $\theta$, with ranges $(-\infty, +\infty)$, $(0, +\infty)$ and $(0,.5)$.
In this family, $E(X) = \mu$, $Var(X) = \sigma^2$, and the median absolute deviation from the mean is $0.5\sigma$.
The kurtosis of $X$ is as follows:
kurtosis $= E(Z^4) = .5^4 * .5 + 1.2^4 * (.5 - \theta) + (0.155/\theta + 1.44)^2 * \theta$.
Within this family,
(i) kurtosis tends to infinity as $\theta \rightarrow 0$.
(ii) the distribution within the "shoulders" (i.e., within the $\mu \pm \sigma$ range) is constant for all values of kurtosis; it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to one interpretation of kurtosis, which states that larger kurtosis implies movement of mass away from the shoulders, simultaneously into the range between the shoulders and into the tails.
(iii) the "peak" of the distribution is also constant for all value of kurtosis; again, it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to the often given but obviously incorrect interpretation that larger kurtosis implies a more "peaked" distribution.
In this family, the central portion of the distribution actually becomes flatter as kurtosis increases, since the probabilities on $\mu \pm 1.2\sigma$ and $\mu \pm 0.5\sigma$ converge to the same value, $0.25$, as the kurtosis increases.
(iv) The median absolute deviation from the mean is constant, $0.5\sigma$, for all values of kurtosis.
