We have defined the Q as the following:
Now I tried to apply that to the parallel tank circuit, I get : $$Q = RC\omega_0 + \frac{R}{\omega L}$$
However in the lecture notes and the textbook the formula used is: $$Q = RC\omega_0 $$
Question is: Why is the inductor ignored in this formula? and does this have something to do with the Q of an inductor being much smaller than that in general of a capacitor ? this is from this answer.
When you tried to apply your ideas to the parallel RLC you used "R" twice mistakenly. Q is either-one of the two terms you calculated but, not both added together.
And, it simply follows (from either term) that at resonance, \$Q = R\sqrt{\dfrac{C}{L}}\$
Q is often defined as \$ Q = {\omega _o \over bandwidth} \$ for a bandpass filter, where \$ \omega _o \$ is the resonant frequency of the filter.
You can construct a simple RLC (parallel LC components) using a low-pass prototype (RC components) and adding an inductor so the LC combination resonates at the desired center frequency. The low-pass prototype determines the bandwidth of the filter, i.e. if you want a 1 kHz bandwidth, create an RC low-pass to give a corner frequency equal to the desired bandwidth of the bandpass filter. See this Stackex answer on how to build a simple series RLC bandpass filter.
Getting back to your question, the bandwidth of the filter is the corner frequency of the low-pass prototype which is \$ f_c = bandwidth = {1 \over {RC}} \$ . Plugging this in to the equation in the first paragraph gives \$ Q = {RC \omega _o} \$ . The value of the inductor is wrapped up in \$ \omega _o = {1 \over \sqrt{LC}} \$ .
This is covered in "Simplified Modern Filter Design", Philip Geffe.
Now I tried to apply that to the parallel tank circuit, I get : $$Q = RC\omega_0 + \frac{R}{\omega L}$$.
Q-factors don't combine this way. For a parallel RLC circuit at resonance the reactance is infinite with a phase angle of zero. But current circulates back and forth between L and C just the same.
For the passive components in parallel, the Q-factors are given as shown in the OP.
$$Q_{L}=\frac{R}{L\omega},Q_{C}=RC\omega$$
These Q-factors are frequency dependent.
If the energy stored in the capacitor (or inductor) is summed over one cycle you get zero. The capacitor and the inductor still provide storage at the same time. They are very active.
The Q for the parallel RLC, \$Q_P\$ is obtained using the geometric mean.
$$Q_{P}=\sqrt{Q_{L}Q_{C}}=\sqrt{\frac{R}{L\omega}RC\omega}=R\sqrt{\frac{C}{L}}$$
Notice that \$Q_P\$ is not frequency dependent.
This is validated with the result from 2nd order analysis of the circuit. Compare the first degree terms of the characteristic polynomial (CP):$$s^2+\frac{1}{RC}s+\frac{1}{LC}$$ with the standard form, $$s^2+\frac{\omega_0}{Q}s+\omega_0^2.$$ Solving for Q will give the same expression.
Notice that the centre frequency \$\omega_0=\omega_N\$, the natural frequency.
However in the lecture notes and the textbook the formula used is:$$Q=RC\omega_0$$
The expression for \$Q_P\$ can be manipulated: $$Q_{P}=R\sqrt{\frac{C}{L}}=R\sqrt{\frac{C^{2}}{LC}}=RC\omega_{0}$$
Question is: Why is the inductor ignored in this formula?
It isn't ignored. It is included with \$\omega_0\$. \$Q_P\$ can also be written: $$Q_{P}=R\sqrt{\frac{C}{L}}=R\sqrt{\frac{LC}{L^2}}=\frac{R}{L\omega_{0}}$$
Extras, but related
The magnitude response at the high and low cutoff frequencies must be equal so \$\frac{H(\omega_h)}{H(\omega_l)}=1\$. Solve for the centre frequency to show that \$\omega_0=\sqrt{\omega_h\omega_l}\$.
The phase at the corner frequency is \$45^0\$. So the tangent, \$\frac{\omega_{0}\omega_{l}}{Q_P\left(\omega_{0}^{2}-\omega_{l}^{2}\right)}=1\$ can be used to show that \$Q_P=\frac{\omega_0}{\omega_h-\omega_l}\$, the centre frequency divided by the bandwidth.
