I have given a similar issue some thought and it seems that the answer to your question is on the negative in general but positive when the input has constant norm. To gain intution consider the simple case of a scalar Gaussian channel with unit noise variance and discrete input with alphabet $\mathcal X$ and probability mass function $p(x)$ for every $x \in \mathcal X$. We then have $$E\[X|Y=y]=\frac{\sum_{x \in \mathcal X}xp(x)e^{-\frac{1}{2}(y-x)^2}}{\sum_{x \in \mathcal X}p(x)e^{-\frac{1}{2}(y-x)^2}}=\frac{d}{dy}\log \( \sum_{x \in \mathcal X}p(x)e^{-\frac{1}{2}x^2 + xy}\)$$ Now, notice that $\log \( \sum_{x \in \mathcal X}p(x)e^{xy}\)$ is the comulant generating function of $X$ at point $y$. Therefore, if $X^2$ is deterministic then the conditional expectation can indeed be expressed as a power series in $y$ with coefficients that are the comulants of $X$. Otherwise, the conditional expectation can still be expanded to a power series in $y$ for fun and profit, but the coefficients are not likely to depend on the cumulants in any simple way, at least as far as I see.

I have given a similar issue some thought and it seems that the answer to your question is on the negative in general but positive when the input has constant norm. To gain intution consider the simple case of a scalar Gaussian channel with unit noise variance and discrete input with alphabet $\mathcal X$ and probability mass function $p(x)$ for every $x \in \mathcal X$. We then have

$$E[X|Y=y]=\frac{\displaystyle\sum_{x \in \mathcal X}xp(x)e^{-\frac{1}{2}(y-x)^2}}{\displaystyle\sum_{x \in \mathcal X}p(x)e^{-\frac{1}{2}(y-x)^2}}=\frac{d}{dy}\log\left( \sum_{x \in \mathcal X}p(x)e^{-\frac{1}{2}x^2 + xy}\right).$$ Now, notice that $\log(\sum_{x \in \mathcal X}p(x)e^{xy})$ is the comulant generating function of $X$ at point $y$. Therefore, if $X^2$ is deterministic then the conditional expectation can indeed be expressed as a power series in $y$ with coefficients that are the comulants of $X$. Otherwise, the conditional expectation can still be expanded to a power series in $y$ for fun and profit, but the coefficients are not likely to depend on the cumulants in any simple way, at least as far as I see.