Solution:

Answer: $\frac{1}{2}$

We first claim that if a $2n$-digit number $x$ has at least eight $0$'s and at least eight $1$'s and the sum of its digits is even, then $x$ is special.

Let $A$ be a set of eight $0$'s and eight $1$'s and let $B$ be the set of all the other digits. We split $B$ arbitrarily into two sets $Y$ and $Z$ of equal size. If $\left|\sum_{y \in Y} y - \sum_{z \in Z} z\right| > 8$, then we swap the biggest element of the set with the bigger sum with the smallest element of the other set. This transposition always decreases the absolute value of the sum: in the worst case, a $9$ from the bigger set is swapped for a $0$ from the smaller set, which changes the difference by at most $18$. Therefore, after a finite number of steps, we will have $\left|\sum_{y \in Y} y - \sum_{z \in Z} z\right| \leq 8$.

Note that this absolute value is even, since the sum of all the digits is even. Without loss of generality, suppose that $\sum_{y \in Y} y - \sum_{z \in Z} z$ is $2k$, where $0 \leq k \leq 4$. If we add $k$ $0$'s and $8-k$ $1$'s to $Y$, and we add the other elements of $A$ to $Z$, then the two sets will balance, so $x$ is special.