As noted in the linked Wikipedia article on the [Fabius function][1], on the interval $I:=[0,1]$ the Fabius function coincides with the cumulative distribution function (cdf) of 
$$\sum_{j=1}^\infty 2^{-j}U_j,$$
where the $U_j$'s are independent random variables uniformly distributed on $I$. So, for each $x\in I$ 
$$F(x)=\lim_{n\to\infty} F_n(x),\tag{1}$$
where $F_n$ is the cdf of $\sum_{j=1}^n 2^{-j}U_j$. 

Next (see e.g. [formula (2.2)][2]), for any real $x$
$\newcommand\vp{\varepsilon}$
\begin{equation}
	F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x})
	=\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\,
	\big(x-c^{(n)}\cdot\vp\big)_+^n,  
\end{equation}
where $\text{vol}_n$ is the Lebesgue measure on $\mathbb R^n$, $H_{n;b,x}:=\{v\in\mathbb R^n\colon b\cdot v\le x\}$, $c^{(n)}:=(c_1,\dots,c_n)$, $c_j:=2^{-j}$, $|\vp|:=\vp_1+\dots+\vp_n$, $\cdot$ denotes the dot product, and $t_+^n:=\max(0,t)^n$. So, for $x\in I$
\begin{equation}
	F_n(x)=\frac{2^{n(n+1)/2}}{n!}\,\sum_{y\in D_{n,x}}(-1)^{s_n(y)}\,
	\big(x-y\big)^n, \tag{2} 
\end{equation}
where $D_{n,x}$ is the set of all dyadic numbers in $[0,x]$ of the form $m2^{-n}$ for integers $m$, and $s_n(y)$ is the sum of the binary digits of $y$.

Formulas (1) and (2) provide the answer to the question. 

----

Some of the differences between (1)--(2) and the formula conjectured in the linked MathSE post are as follows: 

 - In the MathSE post, the main conjectured formula for $F(x)$ is stated only for dyadic numbers $x$, and then extended to all values of $x$ by the continuity of $F$. 
 - The expression in that post for $F(x)$ for dyadic $x$ contains a double summation and a number of $q$-Pochhammer symbols (I have had no experience with those symbols). 

 


  [1]: https://en.wikipedia.org/wiki/Fabius_function
  [2]: https://arxiv.org/abs/1909.06406v1