Here's a counterexample in $n\ge3$ dimensions. With $\lVert\cdot\rVert$ denoting the Euclidean norm, set $$ f(x)=\begin{cases} x_1x_2x_3\lVert x\rVert^{-n-4},&{\rm if\ }x\not=0,\cr 0,&{\rm if\ }x=0. \end{cases} $$ This is harmonic on $x\not=0$, has first and second order derivatives at the origin, but $f$ along with all its derivatives are discontinuous at the origin.
Note that, $f=c\frac{\partial^3(\lVert x\rVert^{-n+2})}{\partial x_1\partial x_2\partial x_3}$ is a third order derivative of a harmonic function, so is harmonic (away from the origin). As $f$ vanishes whenever all but two coordinates are zero, it along with its first order derivatives all vanish on the coordinate axes. So, $f$ along with all its first and second order derivatives vanish at the origin.
Next, here's a counterexample in $n=2$ dimensions. Using $i=\sqrt{-1}$, define $f\colon\mathbb{R}^2\to\mathbb{R}$ by, $$ f(x,y)=\begin{cases} \Re\left[\exp\left(-(x+iy)^{-4}\right)\right],&{\rm if\ }(x,y)\not=0,\cr 0,&{\rm if\ }x=y=0. \end{cases} $$ As the real part of an analytic function, $f$ is harmonic away from the origin. Also, $f$ along with its derivatives to any order are bounded by a multiple of a power of $\lVert x\rVert$ mutiplied by $\exp(-\lVert x\rVert^{-4})$ on any bounded subset of the coordinate axes (outside of the origin). So, $f$ together with all partial derivatives to all orders exist and are zero at the origin. On the other hand, $f$ is not continuous at the origin. Note, this also gives an alternative counterexample to the first one above in $n\ge3$ dimensions by making $f$ independent of $x_3,\ldots,x_n$.