Based on provided data you will be able to comfortably bring the water to boiling point in 1 hour. The energy input required to heat the water alone is significantly less than the available energy and the difference is greater than typical thermal losses.
ie Water_mass x delta_temperature x Water_specific_heat < energy input
As an induction cooker produces temperatures in the base of the cooking vessel which are only slightly higher than the liquid contents (enough to cause thermal transfer) it is possible to insulate the cooking container to reduce thermal losses if desired.
ie If desired and there are no other special factors the pot can be wrapped in a towel or other insulator to reduce thermal losses.
(Remove towel before using pot or electric or gas stove :-) ).
A rolling boil will easily be achieved once the water reaches 100 C as the energy required to increase temperature to boiling point is now all available to cause phase change (aka boiling) while thermal losses will be about the same as during prior heating.
You use a temperature delta of 95 degrees implying cold water at 5 degrees to start.
In many cases water will be somewhat warmer as it tends to be at in-ground temperature (depending on source) but a delta of 95 makes for a close to worst case example.
I'll assume that energy to heat water is ~= 4200 J/kg/K (you use essentially same 4181 )
and that water density is 1 kg/l. In practice both water density and specific heat vary with temperature across the temperature range concerned. A rounded figure of 4.2 kJ/kg/K is accurate enough for for practical purposes given the various other uncertainties such as the electronic grid power to thermal conversion efficiency and thermal losses.
Required: Heat 11 litres of water x 95 C in 1 hour.
Energy = 4200 x kg x delkta_t J = 4200 x 11 x 95 ~= 4.4 MJ
Same as your answer (no great surprise :-) ).
Required electrical input at 85%
= 4.4MJ / 0.84 = 5.2 MJ into heater.
Power to achieve this in one hour = 5.2E6 /(3600s x 1000W) kW
= 1.44 kW
ie if the container had no other losses apart from what was assumed by Wikipedia then a 2000 Watt input unit would achieve the result comfortably (notionally in about 1440/2000 x 60 ~= 43 minutes).
To boil in an hour you can lose (2000-1440)/2000 = 0.28
or ABOUT 25% of your actual heat energy to thermal losses from convection, radiation and conduction. This sounds very achievable, and with an induction cooker you can insulate the pot by eg wrapping it in a towel! - but NOT under the base. (Depending on pot material you can put a thin layer of insulator under the pot an still heat - noting that this is already done as the "cooking surface" is an insulator (probably glass).
Latent heat of vaporization or water (energy required to convert 1 kg from liquid to "steam" is 2260 kJ/kg = 2260 J/g = 2260 J/cc
If 1440 J/s (Watts) was available for the actual boiling process then rate of conversion to steam = Power/(latent heat of vaporisation)
= 1440/2260 = 0.636 cc/second
= 2.29 l/hour
That's a rather small percentage of total volume per second.
However, mentally visualising a 11 litre pot of water on a standard radiant element at full power (ay 1500 Watt) with the water at 100 degrees C - I'm essentially certain that it would boil very vigorously indeed - especially so if it ws possible to insulate the sides of the container, as it probably is in this case.
Additionally, what power rating could I go down to if I were to increase the time to boil by 50%?
As above, if 2000 Watt nominal input gives an ~= 45 minutes to boiling, and the target was < 1 hour, then you'd expect an increase in time to boiling by a factor of 1.5 to 90 minutes to require ABOUT 2000 x 45/90 = 1000 Watts electrical input if thermal losses are able to be scaled down proportionally. With "normal" cooking energy sources adding insulation is problematic but with induction heating adding another towel wrapper and a layer of padding on top is actually feasible. If losses cannot be scaled down then see above calculations for assumed losses and recalculate accordingly.