matrix expression for $e^{iA}Be^{iA}$ in terms of anticommutators?

Let $A,B$ be fixed square matrices of the same size, and define $F(t) = e^{itA}Be^{itA}.$ We then have $$\begin{align} F'(t) &= e^{itA}iABe^{itA} + e^{itA}BiAe^{itA} = e^{itA}(iAB+BiA)e^{itA} = e^{itA}i\{A,B\}e^{itA} \\ F''(t) &= e^{itA}iA(iAB+BiA)e^{itA} + e^{itA}(iAB+BiA)iAe^{itA} = e^{itA}i^2\{A,\{A,B\}\}e^{itA} \\ F'''(t) &= \cdots = i^3e^{itA}\{A,\{A,\{A,B\}\}\}e^{itA} \\ &\vdots \\ F^{(k)}(t) &= i^k e^{itA}(\{A,\bullet\}^kB)e^{itA} \end{align}$$ so $$ F(t) = \sum_{k=0}^{\infty} \frac{t^k}{k!} f^{(k)}(0) = \sum_{k=0}^{\infty} \frac{t^k}{k!} i^k (\{A,\bullet\}^kB) $$ and $$ e^{iA}Be^{iA} = F(1) = \sum_{k=0}^{\infty} \frac{1}{k!} i^k (\{A,\bullet\}^kB). $$

Nice question and nice answers. There is a generalization that admits an even simpler proof that I think it's worth mentioning.

For (square) matrices $X,Y,B$, let me define

$$ B(t):=e^{tX} B e^{tY}. $$

Differentiating with respect to $t$ we see that $B(t)$ satisfies the following differential equation:

\begin{align} \frac{dB}{dt}(t) &= XB(t) + B(t) Y \\ &= \mathcal{K}_{X,Y}(B)& = XZ+ZY, \tag{1} \end{align}

having defined the linear (super) operator $$ \mathcal{K}_{X,Y}(Z) = XZ+ZY $$

Now the solution of $(1)$ with initial condition $B(0)=B$ is $ B(t) = e^{t \mathcal{K}_{X,Y}} B$. We have then

$$ B(t) = e^{tX} B e^{tY} = e^{t \mathcal{K}_{X,Y}} B =\sum_{n=0}^\infty \frac{(t\mathcal{K}_{X,Y})^n}{n!} B . $$

Set $Y=\pm X$ to obtain your formulas with commutators/anticommutators.