Question

how to calculate breakdown voltage of a combination of capacitors?

Answer 1

   Let two capacitors be C1 and C2.   Their breakdown voltages be $V1_{bd}, V2_{bd}$.

   If the two capacitors are connected in parallel, then the smaller of the two breakdown voltages will be the breakdown voltage of the combination.  The reason is that then that capacitor will become a short circuit and will conduct electricity.

   If the two capacitors are connected in series, then the larger of the two capacitances will be the breakdown voltage of the combination.  Because the circuit between the ends of the two capacitor combination, becomes a short circuit only at the larger of the two break down voltages.

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Breakdown voltage of a Capacitor
 
      A capacitance is made of a dielectric medium between two conducting ends/plates etc.  The dielectric could be oil, air, mica etc.   Any medium has a maximum tolerance of some Million Volts / meter  electric field.  Above that electric field strength $E_bd}$, the material conducts electric charges and becomes a conductor.  The distance d  between the two plates/ends of the capacitor  multiplied  by  breakdown electric field strength is the breakdown voltage.
            $V_{bd}=E_{bd}*d$

  For   Air (dielectric inside capacitance) :  E_bd = 2 Million Volts/meter
    In electronics applications, with small voltages, capacitors have the order of picofarads or nano farads.  The width of dielectric medium in between conducting terminals of the capacitors is about  0.01mm 0.1 mm... 

   Hence the breakdown voltage will be:  2 * 10^6 * 0.01 *10^-3 = 20 volts   to  200 volts.

 Oil , mica have breakdown electric fields of the order of 10 times to 100 times.