It will still have self resonance, but the Q will generally be much lower due to the low resistance of the power source 'shorting out' the capacitor (though if the source resistance is high then significant resonance could still occur). Putting the capacitor after the inductor has the opposite effect. Since the source resistance is now in series with the capacitor, a lower resistance will increase Q.
The simplified circuits below show the difference. In the first case the resistor is in parallel with the resonant circuit, so lower resistance reduces Q because the resistor draws more current from it. In the second case it is in series with the resonant circuit, so lower resistance increases Q because it drops less voltage.
simulate this circuit – Schematic created using CircuitLab
But this doesn't mean putting the capacitor in front of the coil is good idea. The main reason for having it after the coil is to attenuate frequencies above the resonant frequency, which it cannot do when placed before the coil.
If you have a very low resistance source that causes significant resonance then you should decrease Q by other means, eg. by adding resistance in series or parallel with the coil. What method is best will depend on the particular parameters of your design. In your example, simply using an inductor with internal resistance of 1Ω is enough to damp the resonance.