Intro by Reid Barton
I think the answer should involve the additivity of variance for independent variables and the central limit theorem. Maybe someone can flesh this out.
Answer
Indeed, the variance is defined to have the additivity property: if r_1 is a random variable with mean m_1 and variance d_1 and r_2 is a random variable with mean m_2 and variance d_2 and these two variables are independent then the new random variable r = r_1+r_2 has the mean m_1+m_2 and variance d_1+d_2.
This will obviously fail for any other function of variance, be it square, cube or something else. Answers that stress convenience are, unfortunately, missing the crucial point.
Moreover, suppose we sum a large number N of independent copies of our random variable r with mean m and variance d. Under mild assumptions, the central limit says the distribution will approach a normal distribution, which by the above has mean Nm and variance Nd. Observe that a normal distribution is completely determined by its mean and variance. We conclude that the only parameters of a distribution that we can observe from the sum of many independent copies of the distribution are the mean and variance.
Now that we established how good it is to square numbers, to get variance, the standard deviation has a very easy explanation: it's the only way to get back from variance to something with the dimension or original set. That is, suppose you numbers are some lengths written in meters. Since the variance is meters squared, you have to take the square root to get something that can be compared with the original set.
(To get back to something in the same units as the original variable, we take the square root of the variance and call it the standard deviation.)
Now, honestly, this not the only way, since you could also, e.g., multiply it by 2. That's why it's called standard deviation — to show that among different numerical constants we've chosen a specific one.