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.