How big is a particular n!?

Yes, Stirling's formula:

\begin{equation} n! \; \sim \; \sqrt{2\pi n} \left(\frac{n}{e}\right)^n. \end{equation}

Another answer in this thread suggests Stirling's approximation: $$n! \approx \sqrt{2\pi n}\biggl(\frac ne\biggr)^n$$

which I think is the right place to start for any answer to this question. However, it may be useful to add that if you take the logarithm of this amount, you get the (approximate) number of digits you need to write down $n!$, which may be of more use:

$$\begin{align} \log n! & \approx \color{darkblue}{n\log n} - n\color{maroon}{\log e} + \color{darkblue}{\frac12 \log n} + \color{darkgreen}{\frac12\log 2\pi}\\ & \approx \color{darkblue}{\biggl(n+\frac12\biggr)\log n} - \color{maroon}{0.43}\cdot n + \color{darkgreen}{0.798} \end{align} $$

($\log$ here is the common, base-10 logarithm function.)

For small $n$ this gives the following values, which I have rounded off to the nearest integer:

$$\begin{array}{rr} n & \text{length of $n!$} \\\hline 1 & 0 \\ 2 & 1 \\ 3 & 1 \\ 4 & 2 \\ 5 & 2 \\ 6 & 3 \\ 7 & 4 \\ 8 & 5 \\ 9 & 6 \\ 10 & 7 \\ 11 & 8 \\ 12 & 9 \\ 13 & 10 \\ 14 & 11 \\ 15 & 13 \\ 16 & 14 \\ 17 & 15 \\ 18 & 16 \\ 19 & 18 \\ 20 & 19 \\ \end{array} $$

This is correct for all $n$ shown except $1$ and $5$, for which it is off by 1.

A cool alternative to Stirling's formula could be the more accurate formula derived by Srinivasa Ramanujan, $$ \ln(n!) \sim n\ln(n) - n + \frac{1}{6}\ln(n(1+4n(1+2n))) + \frac{1}{2}\ln(\pi). $$