estimate the error term in CLT

Stein's method typically gives good Berry-Esseen type bounds for smooth test functions. See Chapter III of Stein's book (entirely viewable in Google Books). For example, specializing to your case of symmetric Bernoulli summands, equation (37) on p. 38 gives $$ \vert \mathbb{E}f(X_m)-\mathbb{E}f(X)\vert \le \frac{2\Vert f' \Vert_\infty}{\sqrt{m}}. $$ For more general summands, there is some simple dependence on the third and fourth moments as well as $\Vert f \Vert_\infty$.

Also, I'm pretty sure that $m^{-1/2}$ is the correct rate here even for Bernoullis, although I can't find a reference for a lower bound at the moment. Why do you expect better?

Your conjecture is correct. Suppose that (say) $f$ has a bounded $5$th derivative. Then
\begin{equation} Ef(X_m) - Ef(X)=-\frac{Ef''''(X)+o(1)}{12m}. \tag{1} \end{equation} Indeed, let \begin{equation} Z_{mj}:=\frac{Z_j}{\sqrt m},\quad Y_{mj}:=\frac{Y_j}{\sqrt m},\quad T_{mk}:=\sum_{j=1}^{k-1}Z_{mj}+\sum_{j=k+1}^m Y_{mj}, \end{equation} where $Z_1,\dots,Z_m,Y_1,\dots,Y_m$ are independent random variables and $Y_j\sim N(0,1)$. Note that $EZ_{mj}^p=EY_{mj}^p$ for $p=0,\dots,3$, $EZ_{mj}^4=\tfrac1{m^2}$, $EY_{mj}^4=\tfrac3{m^2}$. Also, $Z_{mk}$ is independent of $T_{mk}$ for each $k$, and so is $Y_{mk}$. Moreover, $Ef''''(T_{mk})\to Ef''''(X)$ uniformly in $k=1,\dots,m$ (as $m\to\infty$); this should be as easy to show as the convergence $Ef(X_m) \to Ef(X)$, mentioned in the question.

Now one has
\begin{equation} Ef(X_m) - Ef(X)=\sum_{k=1}^m D_k, \tag{2} \end{equation} where
\begin{equation} D_k:=E[f(T_{mk}+Z_{mk})-f(T_{mk}+Y_{mk})] \end{equation} \begin{equation} =\sum_{p=0}^4\frac1{p!}\,Ef^{(p)}(T_{mk})E(Z_{mk}^p-Y_{mk}^p)+O(E(|Z_{mk}|^5+|Y_{mk}|^5)) \end{equation} \begin{equation} =\tfrac1{4!}\,(Ef''''(X)+o(1))(\tfrac1{m^2}-\tfrac3{m^2})+O(m^{-5/2}) \end{equation} by Taylor's expansion, with $o(1)$ uniform in $k=1,\dots,m$. Now $(1)$ follows immediately from $(2)$.