Factorial of 1,e+80

Easiest way to estimate that number is to employ the Stirling approximation: you have

$$ n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$

(this is asymptotic notation), or even more roughly speaking, $ n! \approx e^{n\ln n} $ as order of magnitude. In your case, say $ n = 10^{80} $, you have

$$ n! \approx e^{80\cdot\ln{10}\cdot10^{80}} \approx 10^{10^{81.9}}. $$

(EDIT: I had written 81 instead of 81.9 at the exponent; not only was the rounding incorrect, but especially, rounding exponents needs a lot of care when talking about orders of magnitude!)

By the way, this is also the plausible order of magnitude (in natural units) for the volume of the phase space of the observable universe.

Pictorially speaking, the decimal expansion of this number would need to convert every single particle in the observable universe into a digit in order to be written out.

For an even more pictorial comparison, the age of the universe is thought to be about $ 10^{17} $ seconds. This is far less than the number of digits you would need to write out the above number. Even if you could churn out, say, a billion of digits per seconds, you would still need about $ 10^{82}/10^9 = 10^{73} $ seconds to complete the task, which is a ridiculously prohibitory amount of time.

For the computation time aspect of the question:
Let's assume that a single multiplication takes "constant" time (and that we can perform ~$10^9$ such operations in a second--a reasonable approximation with today's processor).

To compute the factorial of $n$, you must perform $n$ multiplications. Recall that we can perform $10^9$ multiplications per second. So, the runtime of that factorial is: $$\frac{10^{80} \text{ multiplications}}{10^9 \text{ multiplications per second}} = 10^{71}\text{ seconds} \approx 3\times 10^{63} \text{ years}$$

Basically, you'll never see that factorial program terminate.

With the help of wolfram alpha you get an approximation of the number in basis 10 $$ 10^{10^{10^{1.9133}}} $$ Check here other information that should give you an idea of how big the number is.