Cut off frequency in simulation does not match in calculation
Well, first of all it is not very hard to find the transfer function of this circuit. But I will use the result from Okawa Electric Design:
$$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{out}\left(\text{s}\right)}{\text{v}_\text{in}\left(\text{s}\right)}=\frac{\frac{\text{G}}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}{\text{s}^2+\left(\frac{1}{\text{C}_1\text{R}_1}+\frac{1}{\text{C}_1\text{R}_2}+\frac{1-\text{G}}{\text{C}_2\text{R}_2}\right)\text{s}+\frac{1}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}\tag1$$
Where \$\text{G}:=1+\frac{\text{R}_4}{\text{R}_3}\$.
And the circuit is given by:

Using your values, we can see that:
$$\mathscr{H}\left(\text{s}\right)=\frac{1080000000000000000}{381969\text{s}(721497\text{s}+1600000000)+680000000000000000}\tag2$$
Now, we can find the cut-off frequency \$\omega_0\$ by finding:
$$\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\underline{\hat{\mathscr{H}}}\left(\text{j}\omega\right)\right|\tag3$$
This leads to:
$$\omega_0=\frac{200000000 \sqrt{24 \sqrt{145}+1}}{2164491}\approx1573.52\space\text{rad/sec}\tag4$$
In Hertz, this gives:
$$\text{f}_0=\frac{100000000 \sqrt{24 \sqrt{145}+1}}{2164491 \pi }\approx250.433\space\text{Hz}\tag5$$
I found that last result, by using the following Mathematica-code:
In[1]:=Clear[&quot;Global`*&quot;];
G = 1 + (R4/R3);
R1 = 13545;
R2 = 13545;
R3 = 17000;
R4 = 10000;
C1 = 47*10^(-9);
C2 = 47*10^(-9);
s = I*2*Pi*f;
x = ((G)/(C1*C2*R1*
       R2))/(s^2 + ((1/(C1*R1)) + (1/(C1*R2)) + ((1 - G)/(C2*R2)))*
      s + (1/(C1*C2*R1*R2)));
y = FullSimplify[
   Sqrt[ComplexExpand[Re[x]]^2 + ComplexExpand[Im[x]]^2]];
solution = FullSimplify[Solve[{D[y, f] == 0, f &gt; 0}, f]];
solutionForF = f /. solution[[1]];
n = Solve[{y == (1/Sqrt[2])*Limit[y, f -&gt; solutionForF], f &gt; 0}, f];
Flatten[n] &amp;&amp; Flatten[N[n]]

Out[1]={f -&gt; (100000000 Sqrt[1 + 24 Sqrt[145]])/(2164491 \[Pi])} &amp;&amp; {f -&gt; 
   250.433}