We determine--making strong use of the Mathematica code given by user250938 in the answer to this question--the eight atoms of our 256-dimensional Boolean algebra on three sets. Then, we are able to present a table of imposed constraints and their (now partially revised) associated probabilities fully consistent with this framework. This takes the form
$\left( \begin{array}{ccc} \hline Constraint Imposed & Probability & Quasirandom Estimate \\ \hline \hline \_ & 1 & 1.0000000 \\ \text{C} & \frac{8 \pi }{27 \sqrt{3}} & 0.53742158 \\ \neg A\land \neg B & \frac{21}{44} & 0.47726800 \\A & \frac{4702531}{4247100}-\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.50900327 \\ B & \frac{1}{81} \left(27+\sqrt{3} \log \left(97+56 \sqrt{3}\right)\right) & 0.44597788 \\ A\land B & \frac{974539}{1061775}-\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.43224916 \\ A\lor B & \frac{23}{44} & 0.52273200 \\ \neg A\lor \neg B & \frac{1678081}{4247100}-\frac{4 \pi }{27 \sqrt{3}}+\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.56775084 \\ \text{C}\land \neg A\land \neg B & \frac{1678081}{4247100}-\frac{4 \pi }{27 \sqrt{3}}+\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.45591798 \\ \text{C}\land A & \frac{54029}{386100}+\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.079128512 \\ \text{C}\land B & \frac{2}{81} \left(4 \sqrt{3} \pi -21\right) & 0.018903658 \\ \text{C}\land A\land B & \frac{2}{121} & 0.016528575 \\ \text{C}\land (A\lor B) & -\frac{1678081}{4247100}+\frac{4 \pi }{9 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}} & 0.081503595 \\ \text{C}\land (\neg A\lor \neg B) & \frac{8 \pi }{27 \sqrt{3}}-\frac{2}{121} & 0.52089300 \\ \hline \text{C}\land B\land \neg A & \frac{4 \left(242 \sqrt{3} \pi -1311\right)}{9801} & 0.002374589\\ \neg \text{C}\lor B & \frac{13}{27} & 0.48148148 \\ \end{array} \right)$
(The several integer denominators all have prime factorizations with primes no greater than 13--but certainly not the numerators. The prime 97 plays a conspicuous role.)
To obtain these results, we began by estimating the values of the eight atoms--in the indicated order \begin{equation} A \land B \land C, \neg A \land B \land C, A \land \neg B \land C,A \land B \land \neg C, \neg A \land \neg B \land C,\neg A \land B \land \neg C,A \land \neg B \land \neg C,\neg A \land \neg B \land \neg C \end{equation} as--
$\left\{\frac{2984353}{180555569},\frac{428757}{180555569},\frac{11302706}{180555569},\frac{75060766}{180555569},\frac{82318620}{180555569},\frac{2050053}{180555569},\frac{2555632}{180555569},\frac{3854682}{180555569}\right\} \approx \{0.01652872308,0.002374653977,0.06259959780,0.4157211346,0.4559184768,0.01135413885,0.0 1415426848,0.02134900641\}$.
The estimation procedure--starting by generating six-and-a half billion points (triplets in $[0,1]^3$), only approximately one-thirty-sixth of them being further utilized--is the "quasirandom" one of Martin Roberts https://math.stackexchange.com/questions/2231391/how-can-one-generate-an-open-ended-sequence-of-low-discrepancy-points-in-3d
These eight estimated values (summing to 1) are well fitted, we find (using the Solve command), by $\left\{\frac{2}{121},\frac{4 \left(242 \sqrt{3} \pi -1311\right)}{9801},\frac{524119}{4247100}+\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}},\frac{7909}{8775}-\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}},\frac{1678081}{4247100}-\frac{4 \pi }{27 \sqrt{3}}+\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}},-\frac{434}{8775}-\frac{4 \pi }{27 \sqrt{3}}+\frac{\sqrt{3} \log (2)}{\log (81)}+\frac{\cosh ^{-1}(97)}{54 \sqrt{3}},\frac{70064}{1061775}-\frac{4 \pi }{27 \sqrt{3}}+\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}},\frac{87236}{1061775}+\frac{4 \pi }{27 \sqrt{3}}-\frac{\sqrt{3} \log (2)}{\log (81)}-\frac{\cosh ^{-1}(97)}{54 \sqrt{3}}\right\} \approx \{0.01652892562,0.002374589709,0.06259481829,0.4157208527,0.4559237002,0.01135281657,0.0 1415526980,0.02134902704\}$.
To get these formulas yielded by Solve, we first incorporated into the analysis, the three results--$\left\{\frac{8 \pi }{27 \sqrt{3}},\frac{1}{81} \left(27+\sqrt{3} \log \left(97+56 \sqrt{3}\right)\right),\frac{2}{81} \left(4 \sqrt{3} \pi -21\right)\right\}$--having been obtained through symbolic integration. Then, having strong confidence in the previously (tabulated) used values of $\frac{21}{44},\frac{2}{121}$ and $\frac{8 \pi }{27 \sqrt{3}}-\frac{2}{121}$ expressions, we made use of them too.
Since these six values were not fully sufficient for Solve, we additionally employed WolframAlpha--searching over the 256 BooleanFunctions to find simple well-fitting formulas, using the above-given numerically estimated values of the eight atoms. For instance, for BooleanFunction[133,{A,B,C}]=(A && C && B) || (! A && ! C), WolframAlpha suggested $\frac{16}{325}$, fitting the estimated corresponding value to a ratio of 1.00000006615. Also, for BooleanFunction[62,{A,B,C}]=! (A && B) && (A || C || B), the suggestion was $\frac{\sqrt{3} \log (2)}{\log (9)}$, having an analogous ratio of 0.999999807781.
Incorporating as well, these last two results, as well as the previously tabulated $\frac{13}{27}$ for $\neg C \lor B$, proved sufficient to obtain the eight "atomic" formulas.
The ratios of these formulas to the estimated values, given above, are $\{1.000012254,0.9999729358,0.9999236495,0.9999993220,1.000011457,0.9998835421,1.000070743,1.000000966\}$
A somewhatSomewhat interesting observationobservations with regard to the entries of the revised table isare that $\cosh ^{-1}(97)=\sinh ^{-1}\left(56 \sqrt{3}\right)$$\cosh ^{-1}(97)= \log \left(97+56 \sqrt{3}\right)=\sinh ^{-1}\left(56 \sqrt{3}\right)$, so that $\sqrt{3}$ is even more omnipresent.