What's the closest approximation to $\pi$ using the digits $0-9$ only once?
We will use notation $\sqrt{^{n}\;x}\quad$ for $n$ nested square roots (see tehtmi's answer): $$\sqrt{^{n}\;x} = \underbrace{\sqrt{\sqrt{\cdots \sqrt{\sqrt{x}}}}}_{n} = \sqrt[2^n]{x}.$$
Here is one example:
this formula is pandigital one; it uses all mentioned operations excluding exponentiation. $$ \sqrt{\dfrac{8}{\sqrt{^2\;9}}} + \sqrt{\vphantom{\dfrac{8}{8}}^{8}\; \dfrac{\sqrt{^9\; \sqrt{6!} + \sqrt{^{11}\; \left( \sqrt{^8\;4} + \dfrac{1}{\sqrt{^{16}\;2}} + \sqrt{^9 \left(\sqrt{^7\;5!} - \sqrt{^{16}(3!-0!)}\right) } \right) }}}{7} } \\ \approx 3.14159\;26535\;89793\;23846\;26\color{Tan}{610} \approx \pi + 1.766\times 10^{-23}; \tag{P.1}$$
By steps:
$a = \sqrt{^7 \; 5!} - \sqrt{^{16}\;(3!-0!)} \approx 0.03808 59887 98727 14645 $;
$b = \sqrt{^8 \; 4} + \dfrac{1}{\sqrt{^{16}\;2}} + \sqrt{^9\; a} \approx 2.99905 70159 26621 87896$;
$c = \sqrt{6!} + \sqrt{^{11}\;b}\approx 27.83335 21520 97348 95023$;
$d = \dfrac{\sqrt{^9\;c}}{7}\approx 0.14378 82430 20488 46604$;
$\pi \approx \sqrt{\dfrac{8}{\sqrt{^2\;9}}} + \sqrt{^8\; d}$.
Simplified WolframAlpha checking code:
sqrt(8/sqrt(3)) + ( ( sqrt(6!) + (4^(1/256) + 2^(-1/65536) + ( (5!)^(1/128) - 5^(1/65536) )^(1/512) )^(1/2048) )^(1/512) /7 )^(1/256) - pi
How it can be obtained:
Note that when we will use one digit, we can approximate number $\pi$ to $2$ decimal digits: $$ \sqrt{^{15} \; (7!)!} \approx 3.1\color{Tan}{822} \approx \pi + 4.067 \times 10^{-2}; \tag{1} $$
$2$ used digits: $4$ decimal digits of accuracy: $$ \sqrt{\sqrt{5!} - \sqrt{^7\;7!}} \approx 3.141\color{Tan}{349} \approx \pi - 2.435\times 10^{-4}; \tag{2} $$
$3$ used digits: $7$ (and maybe more) decimal digits of accuracy: $$ 2 + \sqrt{\vphantom{\dfrac{1}{1}}^{\:6} \; \dfrac{7!}{\sqrt{^8 \; 9!}}} \approx 3.14159\;2\color{Tan}{576} \approx \pi -7.676 \times 10^{-8}; \tag{3} $$
$4$ used digits: $11$ (and maybe more) decimal digits of accuracy: $$\sqrt{^{5} \; \sqrt{^8\;8!} - \sqrt{^{14}\;7!}} + \sqrt{6 - \sqrt{^{16}\; 3!}} \\ \approx 3.14159\;26535\color{Tan}{682} \approx \pi - 2.157 \times 10^{-11}; \tag{4} $$
$5$ used digits: $13$ (and maybe more) decimal digits of accuracy: $$ \sqrt{\dfrac{8}{\sqrt{^2\;9}}} + \sqrt{\vphantom{\dfrac{8}{8}}^{8}\; \dfrac{\sqrt{^9\; \sqrt{6!} + \sqrt{^{11}\;3}}}{7} } \\ \approx 3.14159\;26535\;89\color{Tan}{835} \approx \pi + 4.178\times 10^{-14}; \tag{5} $$
Now, replacing the digit "$3$" in $(5)$ by appropriate expression constructed of digits $0,1,2,3,4,5$, we get $(P.1)$ which approximates number $\pi$ with accuracy of $>20$ decimal places.
Approximation without $!$ and $\sqrt{\phantom{88}}$ : $$\Large 3 + \dfrac {9^{^{\frac{2}{5\cdot 7} - \left(1+6\right)^{-4}}}} {8} = 3+ \dfrac{9^{\frac{2}{35}-\frac{1}{2401}}}{8} \\ \approx \normalsize 3.1415926535 \color{Tan}{916} \approx \pi +1.875\times 10^{-12}.\tag{P.2}$$
(see Pi Estimation using Integers for more info).
Another approximation (without multiple square roots):
$$ \left(\sqrt{3!}-\sqrt{\sqrt{90-2}-6}\right) \times \sqrt{1+(4+5)\sqrt{8}} + 0\times 7 \\ = \left(\sqrt{6}-\sqrt{\sqrt{88}-6}\right) \times \sqrt{1+9\sqrt{8}} \\ \approx 3.1415926535897\color{Tan}{423} \approx \pi - 5.089\times 10^{-14}.\tag{P.3} $$ (see The Contest Center - Pi for similar examples).
Some approximations can be derived from Pi Approximations.
$\frac{\log (5280^{\sqrt{9}} + 3!! + 4! )}{1 \times \sqrt{67}} \approx 3.14159265358979324$, good for 18 places
If you don't like logs,
$(8\times9 +\frac{52-0!}{73})(\sqrt{-1}^{\sqrt{-4}}) \approx 3.14159266$, good for 8 places of accuracy.
$\sqrt{\sqrt{\frac{2143}{5!!+7}}} \approx 3.141592653$, good for 9 places of accuracy.
$3 \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{20-\frac{4\times8}{9\times5 - 7 - 1}}}}}}} \approx 3.141592652$, good for 9 places of accuracy.
For $e$, I challenge people to beat Sabey's approximation of $(1+9^{-4^{7\times6}})^{3^{2^{85}}}$, which is only accurate to 18457734525360901453873570 decimal digits.
I can't prove this, but I think we may be able to get as close as we want by starting with $1234567890$, taking some number of factorials, and then taking square roots until we get a number less than $\pi^2$. Heuristically we get a random number in the interval $[\pi, \pi^2]$ whose logarithm is uniformly distributed so we ought to be able to get as close as we like to $\pi$ by taking enough factorials.