Micomaco,
Wallace and others have already answered in a completely satisfactory way, so I'll just throw in some extra non-essential detail.
You asked about how the original temp at last scatter is estimated. I've seen a complete derivation of that and it's a fairly involved estimation involving variables like the percent ionization of the gas and the mean free path of the photons.
The last scattering "surface" is actually not a mathematical surface. It has finite thickness. "Recombination" is not an instantaneous event. The redshift at which it occurred, approx z = 1090, is an "around that time" approximation.
The point is that yes as somebody already said the temp is estimated to be about 3000 kelvin. That corresponds to a particle energy of 0.26 eV. But the ionization energy of hydrogen is 13.6 eV. So at 3000 kelvin only a small percentage of the hydrogen is ionized.
That corresponds to a large but finite mean free path. If the universe were not expanding and thinning out and cooling, then a photon would travel only that far on average before getting scattered again. Like in a fog there would not be perfect transparency, you couldn't "see forever", it would be like "visibility one mile" that you hear aircraft pilots say.
The mathematical condition that determines that you get effective transparency at 3000 kelvin is that the mean free path by then is then long enough and the universe is still expanding fast enough, that by the time the photon can expect to hit charge and be scattered again, the gas has thinned out some more and it doesn't get scattered.
Like if medical science were always increasing your life expectancy which at any given time always finite, but if progress were fast enough so you would be effectively immortal because, by the time you reach the original limit, they have improved stuff and you have a new extended expectancy.
So the photon's mean free path is finite, but it is increasing fast enough that most photons make it to infinity.
That means that the estimate of 3000 K depends on a calculation, and slightly somewhat on which model of expansion is used. But it's still pretty reliable---it's robust, not too dependent on details.