Solution:
By symmetry, we only need to consider the $x$-distance, then we can multiply our answer by 2. Let this quantity be $g(n)=a_{n} / 2$.

Divide the $n$th iteration fractal into three meta-columns of equal width. Then the probability that a random point is in the first, second, and third meta-columns is $\frac{3}{8}, \frac{2}{8}$, and $\frac{3}{8}$, respectively. If the two points end up in neighboring meta columns, the expected value of their $x$-distance is simply the width of a meta-column, which is $3^{n-1}$. If they end up in opposite meta-columns (the left and right ones), it is twice this amount, which is $2 \cdot 3^{n-1}$. Finally, if the two points lie in the same meta-column, which happens with probability $\left(\frac{3}{8}\right)^{2}+\left(\frac{2}{8}\right)^{2}+\left(\frac{3}{8}\right)^{2}=\frac{11}{32}$, the expected $x$-distance is just $g(n-1)$. Thus, we have
$$g(n)=3^{n-1}\left(2 \cdot \frac{3}{8} \cdot \frac{2}{8}+2 \cdot \frac{2}{8} \cdot \frac{3}{8}\right)+\left(2 \cdot 3^{n-1}\right)\left(2 \cdot \frac{3}{8} \cdot \frac{3}{8}\right)+\frac{11}{32} g(n-1)=\frac{15}{16} \cdot 3^{n-1}+\frac{11}{32} g(n-1)$$
As $n$ grows, say this is asymptotic to $g(n)=3^{n} C$ for some constant $C$. Then we can write $3^{n} C=\frac{15}{16} \cdot 3^{n-1}+\frac{11}{32} \cdot 3^{n-1} C \Longrightarrow C=\frac{6}{17}$. Our final answer is twice this, which is $\frac{12}{17}$.