As flatulent points out, the voltage at center (where L and C are connected together) gives an indication of the Q value. Be aware that in all resonant circuits the physical mechanism for obtaining the resonant impendance produces one or another extreme situation: In parallel resonance the current circulating in the circuit is multiplied by Q, while in series resonance the voltage is multiplied.
That high voltage can in extreme cases actually cause flashover, and thus shorts. As a result you might even have an internal short if the circuit first had very high Q (and low impedance you wanted), but subsequently something said "bang" due to the voltage stress, and now either the inductor or capacitor might be damaged. Such partial short would lower the Q value, rsulting that the resonant voltage is lowered proportionally. And you measure higher total impedance.
Also, if you think that the resonant impedance would ever be very close to the DC resistance, dream on. It will be substantially more, but should not necessarily be 100 times more as of now! A circuit has all kinds of additional losses: Coil copper AC losses, coil core losses, capacitor's internal losses.
The fact that you do not observe any substantial heating is maybe because you only dissipate about 1/2 watt total (1 A rms, 555 mV). If the components are bulky enough, it is hard to locate where it heats up.