How to calculate the pH of a buffered solution with Henderson Hasselbalch?
The main equilibrium in such a solution is the following:

$\ce{CH3COOH +CH3COO^- &lt;=&gt;CH3COOH + CH3COO^-}$. 

This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:

$\ce{CH3COOH +H_2O &lt;=&gt;CH3COO^- + H_3^+O}$

You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.


  However, doesn't this assume that the system is already at equilibrium
  and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$
  are the concentrations at equilibrium?


Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations. 

In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following: 

$\ce{CH3COOH +CH3COO^- &lt;=&gt;CH3COOH + CH3COO^-}$. 

As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities. 

We also have these equilibria:

$\ce{CH3COOH + H2O- &lt;=&gt;H_3^+O + CH3COO^-}$
$\ce{H2O + CH3COO^- &lt;=&gt;CH3COOH + HO^-}$

But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system). 



If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution electro neutrality. 

Mass conservation:

$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$

Solution electroneutrality:

$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$. 

These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$.