How to calculate the power dissipation of the snubber RC?
You can roughly estimate the power loss in the resistor by calculating the power needed to charge and discharge the capacitor at the switching frequency of the converter: P=CV²f. With 330pF, 16V and 2.1MHz, this results in about 180mW being dissipated in the resistor.
The formula can be derived like this: The energy stored in a capacitor is  Ecap=1/2 CV². For every switching cycle, the snubber capacitor has to be charged and discharged once, which will dissipate the same amount of energy in the resistor. This means that for each switching cycle, the energy in the capacitor is lost twice, which comes out to Ecycle=CV². Multiply this by the switching frequency and you get the power dissipation.
Of course, there will be voltage spikes as well that the snubber is designed to suppress, so the 180mW are only a lower bound for the power dissipation - the resistor is guaranteed to dissipate at least this much. Assuming that the signal across the snubber will be at most 25V peak-to-peak (including transients; this is just an educated guess), this will then result in almost 500mW of dissipation (using the same formula). In practice, the dissipation will most likely be somewhere between these two extremes. As John Birckhead already said, it's best to measure this in circuit and determine the maximum transient voltage that way.
When in doubt, use a resistor with a large power rating (i.e. 1 Watt in your case). Snubber networks can dissipate a surprisingly large amount of power, especially at high switching frequencies.
The maximum voltage across the capacitor is equal to the transient voltage; you'd have to measure it in the actual circuit. If you want to make sure it's less than a certain voltage, add a Zener diode.