The problem is, I feel, in your thinking. The unit impulse does contain all frequencies and therefore, the RMS value of the particular frequency that corresponds with your filter's resonant frequency is infinitely small.
But, you might say that your filter's bandwidth is still quite wide so the energy of the spectrum around your resonance is not infinitely small. However, all the "in-band" frequencies that are "stimulating" your filter are incoherent and you can't expect to see those energies translated to one clear and obvious sinewave.
Regarding any AC analysis on this circuit, the effect of R and Rload are effectively in parallel and will therefore produce the same Q irrespective of whether their parallel value is in the feed position or the shunt position.
Think about the voltage source and its equivalent circuit with R and Rload - forget about L and C for the moment. Try this for size: -
The two circuits are identical and the little_r resistor has changed to be across a current source. This puts it in parallel with any resistor across terminals A and B.
In fact if V1, Rload (or R1) were inside a box and you were not allowed to look inside, you could never know that what was contained was a voltage source in series with a resistor OR a current source in parallel with a resistor - there's no way of telling.
Here's an even more complex scenario: -
Now, if R2 were zero, the equivalent output impedance is the parallel arrangement of the two other resistors. Does that ring a bell?
It's called Norton's theorum - try googling it - also look up Thevenin's theorum - it operates in the reverse: -
Picture stolen from here. Does this make sense now?
So if you agree that Q = \$\omega C R\$ then you know what value to use for R.
Calculating Q for a parallel LC fed by a voltage source via a resistor. Start with the impedance of a pure LC tuned circuit. This is: -
\$\dfrac{sL}{s^2LC+1}\$ then calculate what Vout would be i.e. the transfer function: -
H(s) = \$\dfrac{\dfrac{sL}{s^2LC+1}}{R + \dfrac{sL}{s^2LC+1}}\$
A bit of algebra and this becomes: -
\$\dfrac{\dfrac{s}{CR}}{s^2+\dfrac{1}{LC} +\dfrac{s}{CR}}\$
The bottom line is clearly (to some LOL) recognizable as the denominator in any damped resonant circuit where the various artefacts are: -
See this document and read the bandpass section to confirm. The above picture is an extract.
So \$2\zeta\omega_0 = \dfrac{1}{CR}\$ and, because Q is \$\dfrac{1}{2\zeta}\$, Q = \$CR\omega_0\$
Best Answer
Theoretically it will have no effect on the resonant frequency. The resonant frequency is purely determined by the capacitor having exactly the opposite reactance of the inductor at a particular frequency and the two reactances cancel leaving the series tuned circuit having only resistance at resonance.
However, with lower values of resistance the peak shape of the resonance will change but the centre point of the peak will remain as previous.