Why is the energy of quantum harmonic oscillator independent of its amplitude? The energy of a harmonic oscillator with amplitude $A$, frequency $\omega$, and mass $m$ is 
$$E=\frac 12 m \omega^2A^2 \, .$$
It is intuitive to think that the energy depends on the amplitude because more the amplitude means that the oscillator has more energy, and similarly if the angular frequency is high even then the energy will be more.
Now let's consider a quantum harmonic oscillator (QHO).
The energy is
$$E=\left( n+ \frac 12 \right ) h\nu \, .$$
No amplitude term is there! This is odd because, even  if you argue that we are dealing in microscopic domain, we all can agree to the fact that, in general for any mass oscillating under some force, if we have more energy then the oscillator will move farther from its mean position and therefore will have more amplitude.
The relation of energy of QHO can't be wrong, where else the above conception of energy for an oscillator also doesn't seems to be wrong.
 A: If you want to understand the classical limit of a harmonic oscillator, it is more meaningful to consider coherent states $$|\alpha \rangle = e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n \rangle$$ for $\alpha$ an arbitrary complex number. Such states satisfy $a |\alpha \rangle  = \alpha |\alpha \rangle$. These states all saturate the Heisenberg uncertainty relation, and (unlike the higher excited states of the harmonic oscillator) do have associated dynamics, namely $e^{-i H t/\hbar} |\alpha \rangle = |\alpha e^{i \omega t} \rangle$ up to an overall phase.
One advantage to using a coherent state is that we can easily define the amplitude of the oscillation. The position and momentum expectation values are just $\langle x \rangle = \sqrt{\frac{2 \hbar}{m \omega}} \text{Re} \alpha$ and  $\langle p \rangle = \sqrt{2 m \omega \hbar} \text{Im} \alpha$ where $\text{Re} \alpha$ and $\text{Im} \alpha$ are the real and imaginary parts of $\alpha$, and both are normally distributed about their mean values. Restricting to real $\alpha$ means the oscillator is maximally displaced (as $\langle p \rangle = 0$), and so we would associate the amplitude $A = \sqrt{\frac{2 \hbar}{m \omega}} |\alpha|$ (in fact this remains correct if $\alpha$ is not real). The coherent state is not an energy eigenstate, but $\langle E \rangle = (|\alpha|^2+\frac12)\hbar \omega$. Then we see that $\langle E \rangle = \frac12 m \omega^2 A^2 + \frac12 \hbar \omega$. The first term matches the expression you have in the classical case. The second term is the ground state energy of the quantum harmonic oscillator. It is $0$ in the classical limit $\hbar \rightarrow 0$, and it can also be consistently removed if you choose to by just shifting the potential down by a constant $\frac12 \hbar \omega$.
While one can attempt to recover the classical limit from the higher energy eigenstates rather than from coherent states, it is quite a bit less satisfying to do so, as the high energy eigenstates describe a particle that is delocalized on a macroscopic scale (with similar issues in momentum space) and which does not evolve in time, while a classical particle in a harmonic oscillator should be localized at a scale much smaller than the length of its oscillation and should oscillate with characteristic angular frequency $\omega$ (which is not the case in a single eigenstate, only in a superposition of multiple energy eigenstates). Since energy eigenstates only depend on a single parameter $n$, even for large $n$ where the spacing between adjacent energy levels is negligible the energy eigenstates can not describe a single classical state (which depends on both initial position and momentum), only the time-averaged phase space density of the state, whereas since $\alpha$ is complex (2 real parameters) we have a one-to-one correspondence between classical states and coherent states as we saw above.
A: I ask students in my modern physics classes to deal with exactly this question, but I take the approach of having them find the physical extent of the motion in terms of a probability distribution function for both the classical and quantum cases.
The text I used most recently reads (in part):

Comparison of spatial probability distribution of a one-dimensional harmonic oscillator between the classical and quantum cases. 
A quantum mechanical solution to the harmonic oscillator (mass
    of a spring) problem is developed in Townsend in section 4.3
    which gives
    \begin{align*}
      \psi_n(x) &= A_n H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right) 
               \exp \left(-\frac{m\omega}{2\hbar}x^2\right) \,,
  \end{align*}
    where $A_n$ is a normalizing factor; $\omega = \sqrt{k/m}$ is
    the angular frequency of the oscillator; $m$ is the mass; and
    $H_n$ indicates the $n$th Hermite polynomial. These polynomials
    are tabulated in many places including Wikipedia.
  
  
*
  
*Plot the normalized probability distribution
  $P(x) = \psi_n^*(x) \psi_n(x)$ for three cases we care about
  ($n \in \{0, 1, 4\}$). Use $m = \omega = 1$
  
*Derive an expression for the normalized classical distribution in terms of position and total energy. (This is a purely classical computation; no wavefunctions are involved!) 
  
*Using the fact that the $n$th solution to the quantum
  harmonic oscillator has energy
  $E_n = \left(n + \frac{1}{2}\right)\hbar\omega$, plot the classical
  probability distribution for the energy for each of the plots
  you did for the quantum cases.
  

And later I give them my computer plotted solutions for the cases I requested and a higher $n$ case:

The solid blue curves represent the classical PDF while the dashed red curves represent the quantum probability.
As you can see the spacial extent of the quantum and classical wave-functions track each other pretty well (there is, of course, some quantum penetration into the classical forbidden zone, but that is to be expected).
In the limit of high $n$ (i.e. macroscopic energy) these two cases approach each other very closely as expected from the correspondence principle.
A: You can calculate the variance of the position coordinate, $\sigma_x^2$, for a general eigenstate of the energy $\psi_n$ to be
$$\sigma_x^2=\frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right) \, .$$
We can replace the $n$ dependance by energy dependence using the relation
$$E_n = \hbar\omega\left( n+\frac{1}{2} \right)$$
and we get
$$\sigma_x^2=\frac{\hbar}{m\omega}\frac{E_n}{\hbar\omega} \, .$$
Rearranging we get
$$E_n=m\omega^2\sigma_x^2 \, .$$
Remembering that for the classical case $$\sigma_x^2=\frac{1}{2}A^2$$ we retrieve the original relation in the quantum case as well.
Check the derivation at this link
A: Let's denote $A$ as $x$ while addressing this, as people usually do with the quantum simple harmonic oscillator (well, that or $q$ anyway). The full energy has a term you forgot due to the momentum, viz. $E=\frac{m\omega^2 x^2}{2}+\frac{p^2}{2m}$. In quantum mechanics, the states of definite energy don't have definite values for $x$ or $p$ (or even $x^2$ or $p^2$), although on average half the energy is due to each term. So the amplitude $x$ is a random variable; in fact, in the ground state it has a Normal distribution. (In higher-energy states, we must multiply the Gaussian pdf by a polynomial function.)
