My understanding so far is the (unbiased) sample variance is modified by Bessel's correction to account for the fact that the first observation of $ (X-\overline X)$ in a given sample X, will always equal zero (is not "free to vary"), as i.a. the mean of the first observation and the first observation will always be the same value. So the number of deviations being averaged is n-1 instead of n.
$$ s^2= \frac{∑(X-\overline X)^2}{\rm n-1} $$
When I look at the formula for the sample mean absolute deviation, however, it is:
$$ Mean\:Absolute\:Deviation = \frac{∑|X-\overline X|}{\rm n} $$
Is it also possible to have an unbiased estimator here using $n-1$ and why/why not?
Source for equations: Salvatore and Reagle, Statistics and Econometrics Second Edition - Eqs. 2.6b and 2.8b
For the mean absolute deviation, the correction $n-1$ does not give an unbiased estimator in general. However, it does works in some cases.
Normal distribution For example, take $X$ standard normally distributed. Then, $E|X-EX|=E|X|=\sqrt{2/\pi}$. Moreover, $$ X_1-\bar{X}=\frac{n-1}{n}X_1-\frac{1}{n}\sum_{i=2}^n X_i=:Y $$ This is random variable $Y$ has a normal distibution with mean zero, and variance: $$ Var[Y]=\left(\frac{n-1}{n}\right)^2+\frac{1}{n} $$ So, $Y$ has the same distribution as $\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}X_1$. We calculate: $$ E\sum_{i=1}^n|X_i-\bar{X}|=nE|Y|=nE\left|\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}X_1\right| =n\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}\sqrt{\frac{2}{\pi}} $$ $$ =\sqrt{(n-1)^2+n}\sqrt{\frac{2}{\pi}} $$ So, an unbiased estimator for $E|X|$ is: $$ \frac{1}{\sqrt{(n-1)^2+n}}\sum_{i=1}^n|X_i-\bar{X}| $$
Bernoulli distribution However, it is true for some special cases. If $X$ is $0.5$-Bernoulli, it works. We want to estimate $E|X-EX|$. We have $E|X-EX|=E|X-0.5|=0.5*(0.5+0.5)=0.5$. So, our estimator should have the same expected value. Consider $E|X_1-\bar{X}|$. We can write: $$ E|X_1-\bar{X}|=E\left[1-\frac{1}{n}\left(1+\sum_{i=2}^nX_i\right)\Bigg|X_1=1\right]\frac{1}{2}+ E\left[\frac{1}{n}\sum_{i=2}^nX_i\Bigg|X_1=0\right]\frac{1}{2} $$ $$ =\frac{n-1}{2n}-E\left[\frac{1}{n}\sum_{i=2}^nX_i\right]\frac{1}{2}+E\left[\frac{1}{n}\sum_{i=2}^nX_i\right]\frac{1}{2}=\frac{n-1}{2n} $$ From here, it follows that $\frac{1}{n-1}\sum_{i=1}^n|X_i-\bar{X}|$ has expectation $0.5$, hence it is an unbiased estimator for $E|X-EX|$.