Why $\sqrt{n}$ in the definition of asymptotic normality? A sequence of estimators $U_n$ for a parameter $\theta$ is asymptotically normal if $\sqrt{n}(U_n - \theta) \to N(0,v)$. (source) We then call $v$ the asymptotic variance of $U_n$. If this variance is equal to the Cramer-Rao bound, we say the estimator/sequence is asymptotically efficient.
Question: Why do we use $\sqrt{n}$ in particular?
I know that for the sample mean, $Var(\bar{X}) = \frac{\sigma^2}{n}$ and so this choice normalizes it. But since the definitions above apply to more than the sample mean, why do we still choose to normalize by $\sqrt{n}$.
 A: You were on the right track with a sample mean variance intuition. Re-arrange the condition:
$$\sqrt{n}(U_n - \theta) \to N(0,v)$$
$$(U_n - \theta) \to \frac{N(0,v)}{\sqrt{n}}$$
$$U_n  \to N(\theta,\frac{v}{n})$$
The last equation is informal. However, it's in some way more intuitive: you say that the  deviation of $U_n$ from $\theta$ is becoming more like a normal distribution when $n$ increases. The variance is shrinking, but the shape becomes closer to normal distribution.
In math they don't define the convergence to the changing right hand side ($n$ is varying). That's why the same idea is expressed as the original condition, that you gave. In which the right hand side is fixed, and the left hand side converges to it.
A: We don't get to choose here. The "normalizing" factor, in essence is a "variance-stabilizing to something finite" factor, so as for the expression not to go to zero or to infinity as sample size goes to infinity, but to maintain a distribution at the limit.
So it has to be whatever it has to be in each case. Of course it is interesting that in many cases it emerges that it has to be $\sqrt n$. (but see also @whuber's comment below).
A standard example where the normalizing factor has to be  $n$, rather than $\sqrt n$ is when we have a model
$$y_t = \beta y_{t-1} + u_t, \;\;  y_0 = 0,\; t=1,...,T$$
with $u_t$ white noise, and we estimate the unknown $\beta$  by Ordinary Least Squares.
If it so happens that the true value of the coefficient is $|\beta|<1$, then the the OLS estimator is consistent and converges at the usual $\sqrt n$ rate.  
But if instead the true value is $\beta=1$ (i.e we have in reality a pure random walk), then the OLS estimator is consistent but will converge "faster", at rate $n$ (this is sometimes called a "superconsistent" estimator -since, I guess, so many estimators converge at rate $\sqrt n$).
In this case, to obtain its (non-normal) asymptotic distribution, we have to scale $(\hat \beta - \beta)$ by $n$ (if we scale only by $\sqrt n$ the expression will go to zero). Hamilton ch 17 has the details. 
