From How Big Is A Lightning Bolt? we see that a lightining bolt is “an inch wide and five miles long”, and at “50,000 °F”. So in useful units, approximately 3 cm diameter, 8 kilometer long, 28000 K hot.
If we consider that the heat is mostly due to black body radiation (for a perfect black body with an emissivity of $\epsilon = 1$), then the power will be given by the Stef-Boltzmann law:
$$P = A \epsilon \sigma T^4$$
The area, $A$ of the lightning bolt (a cylinder, of course) is given by
$$A= 2 \pi\times(3 \text{ cm})\times 8 \text{ km} \sim 1500 \text{ m}^2$$
And so,
$$5.2 \times 10^{13}\; \text { Watts of power.}$$
Lets say, it lasts 10 miliseconds, so its around $\sim 5 \times10^{11}$ J.
Now to calculate it the amount of photons properly, you would have to consider the spectrum of the black body radiation, and convert the energy density to number of photons using Planks law. I will just use the rule of thumb that "1 Watt of monochromatic visible light is approx $10^{18}$ photons per second".
And so, it would be around:
$$\sim 10^{29}\ \text{ photons.}$$
