How much light is there in space and how heavy is it?

Actually, on a dark night, the fraction of the sky that is light is pretty negligible. That's what it means to be a dark night ;-)

It's actually not hard to get an estimate of the density of light in the universe. Let's say that "light" includes photons of all wavelengths (not just visible light) for simplicity. A straightforward way to do it is to point a wide-spectrum telescope at the night sky and see how much fast it collects energy. (You have to point it away from the sun and other nearby sources like the galactic disc, because these sources emit a large amount of energy at Earth, which is not representative of the universe as a whole.) This was done with the COBE and WMAP satellites (and more recently Planck, with essentially the same result). They found that, if you eliminate contributions from a few specific nearby sources, the radiation in the universe follows a blackbody spectrum with a temperature of $2.73text{ K}$. You can calculate the energy density of such a blackbody like this:

$$u = frac{4sigma T^4}{c} = 4.2times 10^{-14} mathrm{frac{J}{m^3}}$$

This is just four thousandths of a percent of the critical energy density, which is

$$Omega = frac{3 c^2 H^2}{8pi G} = 8.6times 10^{-10} mathrm{frac{J}{m^3}}$$

On the other hand, the density of normal matter (atoms) is estimated to be about 4.6% of the critical density - four orders of magnitude higher. So the photon energy density is completely insignificant compared to the density of baryons, dark matter, or dark energy, and that means it has a negligible gravitational effect on the cosmological evolution of the universe.

Of course, in the vicinity of a star, the photon energy density is much higher because of the star's higher blackbody temperature. Let's take the sun, for example. The sun has a surface temperature of $5778text{ K}$, which means the intensity of radiation it emits is

$$I = frac{P}{A} = sigma T^4 = 6.31times 10^7 mathrm{frac{W}{m^2}}$$

This corresponds to an energy density in the vicinity of the sun's surface of

$$u_text{rad} = frac{I}{c} = 0.211 mathrm{frac{J}{m^3}}$$

However, the energy density of the matter that constitutes the sun is

$$u_text{matter} = rho_odot c^2 = 1.2times 10^{20} mathrm{frac{J}{m^3}}$$

So again, the gravitational effect of the photons, even on a local scale, is completely negligible.

Incidentally, this would not have been the case in the early universe, when the photons had much higher energy and thus their energy density relative to matter was much higher.