Let me first begin with an elementary example, taking $X=[0,1]^2$ in $\mathbb R^2$. It is then easy to see directly that the wave-front-set of $\mathbb 1_X$ is everywhere the conormal bundle except at the four corners $c_j$:
$$
WF\mathbb 1_X=\cup_{1\le j\le 4}(c_j;\mathbb R^2\backslash {(0,0)})\cup\text{conormal bundle at the smooth points}.
$$
To prove this, it is enough to check the wave-front-set of $H(x_1)H(x_2)$ with $H=\mathbb 1_{[0,+\infty)}$ which amounts to check the wave-front-set of $\delta_0(x_1)\otimes\delta_0(x_2)$ which contains at $(0,0)$ every direction since the Fourier transform is 1.

There are of course (much) more refined results. If you take a look at Lars Hörmander's ALPDO first volume, Springer Grundlehren 256 on page 300, you will find the definition of the normal set to any closed subset of $\mathbb R^n$ and Theorem 8.5.6' asserts that the analytic wave-front-set of a distribution $u$ contains the normal set to the support.

For the example you give, I believe that the $C^\infty$ wave-front-set is simply the product of the boundary with $\mathbb R^2\backslash {(0,0)}$. However for $s$ real, we may define the $H^s$ wave-front-set and I guess that the Hausdorff dimension of the fractal set you consider should be linked to the $H^s$ wave-front-set, in the sense that it should be trivial at some threshold $s_0$ and that $s_0$ should be linked to the Hausdorff dimension. The paper MR1277392, by Falcone, is addressing a related question.