In my experience, for most topologies the number of output capacitors required in a design is determined by the ripple current and not by the capacitance, because when you crunch the numbers, you'll find that the minimum capacitance you need to keep the output in regulation will be smaller than the ESR you need to keep the ripple voltage within reasonable limits (usually 1% of the DC output).
This is engineering, so it's always better to calculate rather than assume. Since you know the maximum load, the expected voltage on the capacitor and the on-time of the primary-side flyback switch (which is the period when no energy is transferred to the capacitor and it discharges to power the load) you can calculate how much the capacitor will discharge and determine what size of capacitor you need to keep reasonable regulation. Then, since you know the ripple current, you can calculate how many capacitors you need to keep the ripple voltage reasonable. Go with whichever calculation is larger.
Example
\$ \Delta V = \dfrac{I \times \Delta t}{C} \$
That is to say, a \$ 100 \mu F\$ capacitor being discharged by a 10A load over \$1 \mu s\$ will lose
\$ \Delta V = \dfrac{10A \times 1 \mu s }{100 \mu F} = 0.1V\$
Let's assume that the ESR of this capacitor is \$100 m \Omega\$ and the peak-to-peak ripple current is 4A (not unheard of for a flyback converter).
\$ \Delta V_{ripple} = \Delta I \times R_{ESR} \$
\$ \Delta V_{ripple} = 4A \times 100 m \Omega = 0.4V \$
SO: If 400mV ripple is too high, put more capacitors in parallel until the ESR is low enough to achieve the ripple you need.
As Olin commented, the stability of the converter is also a consideration when choosing output filter components, especially with flyback converters that operate in CCM. [If none of what I've taking about makes sense, you should seek counsel from someone experienced in power supply design.]
The max current limit on the controller is 2.5A, so the inductor is undersized.
If you try to draw more than the saturation limit of the inductor, the inductance will decrease and the inductor current will rise rapidly and the part will go into cycle by cycle current limit until the current demand goes below the current limit.
You could easily overheat the inductor that way, so it would be better to find an inductor that doesn't saturate before 2.5A.
It would certainly cause the output voltage to drop when it happens.
Best Answer
If you can reasonably guess that something is broken due to electrical overstress, the most likely candidate is the integrated circuit.
It is typically the power switches and control circuits that break. The most common mechanisms are over-heating and over-voltage breakdown. And especially considering your previous question, which suggests that you had \$V_{IN}\$ overshoot, over-voltage breakdown is quite likely. When semiconductors experience over-voltage breakdown, they suddenly absorb a lot of power (often in a snapping, positive-feedback fashion), which causes damage.
As for the inductor, essentially the only way you could electrically break it would be to over-heat it until it melts. The solder will melt long before that happens.
I suggest you swap out the IC and see if it works again.