Parallel resonant circuit depending on voltage
For a given applied voltage (\$V_A\$/\$X_L\$) across a parallel resonant LC circuit that is tuned to resonate, the current that flows in the inductor is \$V_A\$/\$X_L\$ and the current that flows in the capacitor is \$V_A\$/\$X_C\$ AND, importantly these two currents are identical in magnitude at resonance.

However they are opposite in sign hence the net current taken from \$V_A\$ is zero amps (after the initial transient has settled down). But, there are still  currents circulating between L and C of equal magnitude.


  I know that bigger L causes bigger Q, but besides from that, will the
  same current flow if I use a really small capacitor (speaking about
  ideal elements now) and a really big inductor at resonant frequency


No, the value of current (although not taken from the supply at resonance) is determined by that supply and either \$X_L\$ or \$X_C\$.


  What are the differences when we change the values of capacitance and
  inductance while still maintaining the same resonant frequency
  (besides Q factor)?


If you double L and halve C to achieve the same resonant frequency, then the circulating current halves. If you halve L and double C then the circulating current doubles.