Think of it this way:
The air around the coil itself has capacitance.
Therefore, the high voltage is attempting to charge that capacitance.
When the voltage is low, it is charged by induction (proximity to the coil).
However, when the voltage exceeds the breakdown field, the air becomes conductive and electrons and ions surge into that space until the electric field strength is below breakdown. Once this is done, the capacitance seen by the coil is larger, because it is conducting into a larger volume than before.
The equivalent circuit is a resistor in series with a capacitor. The capacitor represents the amount of capacitance gained by the increase in size or extent; the resistor represents the voltage drop across the spark.
This therefore loads the coil, reducing its resonant frequency and Q somewhat.
The Q factor of a spark discharge is evidently fairly high, so that a coil driven at high power does not detune very much in the process.
A large top load (high initial capacitance) helps, by "swamping" the change in capacitance. (That is, if we make C larger to begin with, then if the spark causes some increase \$\Delta C\$, it is a smaller percentage change in C.) Also, by being physically larger, the top load acts to couple to the spark, i.e., the spark's capacitance to its surroundings is partly taken up by the top load itself, and so the spark-to-ground capacitance is smaller.
At very high voltages, streamers leave the vicinity of the top load, or strike ground, and both C and Q can go to crap again; it's a matter of degree with respect to power level, and the type of design used (a fixed-frequency SSTC won't tolerate detuning, but a PLL tracking, or self oscillating, type will).
Tim