I have heard on numerous occasions that the Standard Deviation of a data set is meant to represent "the average spread of the data around the mean", and if that's the case why don't we use this formula instead which yields a much more precise representation of the average spread around the mean: $$\frac{\sum_{i=1}^n \Bigl(+\sqrt{(x_i-\overline{x})^2}\Bigl)}{n}$$
where $n$ is the total number of data values , and the $+\sqrt{}$ symbolises taking explicitly the positive square root of $(x_i-\overline{x})^2$
Take the data set $(1,2,2,2,5,5,7,8)$ for example. My formula produces the value 2.25, whereas the traditional Standard Deviation formula, $s=\sqrt{\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}}$ produces the value $2.619$ to 3.d.p.
I am aware that simplifying the expression inside of the sum using laws of indices results in $\Bigl(+(x_i-\overline{x})\Bigl)$ which would result in the sum being equal to zero. See my proof of this result here.
My question is, is the reason I just stated, why this forumla is not used in place of the Standard Deviation, or can this formula be interpreted in the way I intend it to, and if so why is it not used?