I found a different number for $C_L$, please check your calculation:
$C_L=\frac{2 m g}{\rho V^2 S}$
substitute:
$
m = 50000 \textrm{ kg}\\
g = 9.81 \textrm{ m}/\textrm{s}^2\\
V = 462 \textrm{ KTAS} = 237.67\textrm{ m/s}\\
S = 122.6 \textrm{ m}^2\\
\rho = 0.475 \textrm{ kg}/\textrm{m}^3
$
This will give:
$C_L = 0.298$
Note that this lift coefficient is for the total aircraft. It includes the lift of the wing, the fuselage lift and also the negative lift of the vertical stabilizer. There is also a small vertical component of thrust that is neglected.
Now that we have an estimate of the lift coefficient, we can estimate the angle of attack. Normally the lift coefficient is assumed to vary linearly with the angle of attack:
$C_L = \frac{\textrm{d}C_L}{\textrm{d}\alpha}\cdot (\alpha-\alpha_0)$
$\alpha_0$ is the zero-lift angle of attack, the angle of attack at which the wing does not generate any lift. For symmetric airfoils, $\alpha_0 = 0$, for chambered airfoils $\alpha_0 < 0$.
I don't know what that angle is for the A320, it will be difficult to obtain. Let's assume for now it is -1.2 degrees.
For infinite long wings in incompressible flow the lift slope $\frac{\textrm{d}C_L}{\textrm{d}\alpha}$ is $2\pi$.
$(\alpha-\alpha_0) = \frac{C_L}{2\pi} =0.0474 \textrm{ rad} \approx 2.72^\circ$
This would result in an angle of attack of $\alpha \approx 1.52^\circ$
Because the wing of the Airbus A320 is not infinitely long but has a span of about 30 meters (excluding the fuselage) we need to correct for that. The reason we need to correct for the finiteness of the wing is that the circulation will cause the local angle of attack of the wing to be lower than the free stream angle of attack. The effective angle of attack $\alpha_{eff} = \alpha - \alpha _i$
The induced angle of attack $\alpha_i$ is given by:
$\alpha_i = \frac{C_L}{\pi AR} = \frac{C_LS}{\pi b^2}$
substitute:
$
S = 122.6 \textrm{ m}^2\\b = 30 \textrm{ m}
$
gives $\alpha_i = 0.0129 \textrm{ rad} \approx 0.74^\circ$
Adding this to our earlier angle of attack results in:
$ 1.52 + 0.74 = 2.26^\circ$
This may not be very accurate as
a) the zero-lift angle of attack may be very different,
b) the lift slope may be flatter due to compressibility effects
