Why is variance squared? 
The mean absolute deviation is:


$$\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}$$


The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$


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*So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value.

*Why isn't it $|x_i-\bar x|^2$, then? Squaring just enlarges, why do we need to do this?

A similar question is here, but mine is a little different.
Thanks.

The long and the short is that the squared deviation has a unique, easily obtainable minimizer (the arithmetic mean), and an inherent connection to the normal distribution. The absolute deviation, on the other hand, can admit multiple non-unique, potentially laborious to obtain minimizers (medians). For a simple illustration of this, observe that the set $\{0,1\}$ admits for a value $x$ the total absolute deviation ($L_1$ norm) $$|x-0|+|x-1|=\begin{cases}1-2x,&x\le0
\\1,&0<x\le1
\\2x-1,&1<x\end{cases}$$
which can be seen to be a piecewise linear/constant function minimized to $1$ by all $x$ in $[0,1]$. Instances with more points may be even more pathological and not admit a simple method of optimization. On the other hand, the total squared deviation ($L_2$ norm) of the same set would be $(x-0)^2+(x-1)^2=2x^{2}-2x+1$, a quadratic function with a unique minimizer of $x=0.5$, easily obtainable by setting its derivative to zero.
The connection of squared deviations to the normal distribution is highly attractive, first for the distribution's ubiquitous applicability to real world phenomena (hence the name), for instance, for dispersed measurements taken from populations or for errors in measurements. Second, the connection is attractive due to the normal distribution's enormously convenient theoretical properties, for instance, since normal distributions are symmetric about their means, have easily obtainable centers and dispersions, are closed under summation, and so on. Furthermore, from a practical point of view, there is extensive theoretical groundwork already established for the normal distribution, which is opportune to lean on.
These characteristics can ultimately be seen as consequences of the various convenient mathematical properties of $x^2$ lacked by $|x|$, e.g. differentiability everywhere (facilitating minimization), that the set of quadratic functions are closed under summation (the sum of two quadratics is another quadratic), and so on.
So this is not to say that absolute deviations are not used or less applicable than squared deviations. On the contrary. Instead, they are, in many relevant ways, less convenient to apply.
See also, Stats SE Q118.
