Is there some mathematical properties of the distribution which is the output of a probabilistic polynomial-time Turing machine?
Given a probabilistic algorithm $A$, define $f^A(x)=\Pr[A(x)=1]$. To make the question interesting, you have to ask whether $f^A$ (with polynomial $A$) can be close to $g$, for infinitely many values of $n$. Otherwise, as Raphael mentioned, you can pretty much obtain any $g:\Sigma^n\rightarrow [0,1]$, while keeping $A$ polynomial (there is no asymptotics). Consider the problem of how $dist_n(f^A,g)=\sum\limits_{x\in S^n}\left|g(x)-f^A(x)\right|$ grows as a function of $n$. Obviously $dist_n(f,g)\le 2^n$, and a random guessing $A$ achieves $dist_n(f^A,g)\le 2^{n-1}$, can we find a probabilistic polynomial algorithm $A$ such that $dist_n$ grows polynomialy?

Let $L\subseteq\Sigma^*$ be some language, and $f^L:\Sigma^*\rightarrow \{0,1\}$ be its indicator function, i.e. $f^L(x)=1\iff x\in L$. 
If for large enough $n$ it holds that $\sup\limits_{x\in\Sigma^n}\left|f^L(x)-f^A(x)\right|\le\frac{1}{2}-\epsilon$, then $\mathsf{L\in BPP}$ (simulate $A$ several times and take the majority vote, this works for $\epsilon=n^{-d}$). If for large enough $n$ you have $\sum\limits_{x\in\Sigma^n}\left|f^L(x)-f^A(x)\right|\le poly(n)$, then $\mathsf{L\in P/Poly}$, since your advice can contain information about all strings $x\in\Sigma^n$ on which $f^L,f^A$ differ by more than (say) $\frac{1}{3}$, and on the rest of the strings you can take the majority vote as above.

Thus, to answer your question, if $g$ is an indicator function for a hard language, you can expect (under some hardness assumptions) that there is no probabilistic polynomial time algorithm $A$ for which $Dist_n(f^A,g)$ is polynomial in $n$.