EDIT: *added $n=12$ and $13$ and improved the source code by saving $k!$ terms for $k$ consecutive random strikes.*

The Thue-Morse sequence isn't a solution when $n=8$, but it gets close. Here are the fairest sequences with $v_0\ge v_1$ for small $n = $ $1,$ $2,$ $\dots,$ $13$, according to an exhaustive brute-force search, where $v_b$ denotes the expected score for the player choosing when the binary digit is $b$.

$$\begin{array}{rcccc}
n&\text{fairest with }v_0\ge v_1&v_0&v_1&v_0-v_1\\ \hline
2&10&5/3&3/2&1/6\\
3&001&5/2&7/3&1/6\\
4&0110&10/3&16/5&2/15\\
5&11010&62/15&33/8&1/120\\
\\
6&000011&5&5&0\\
 &001110&5&5&0\\
 &110001&5&5&0\\
 &111100&5&5&0\\
\\
7&1000101&377/64&88/15&23/960\\
\\
8&00110110&61/9&27/4&1/36\\
 &10000011&61/9&27/4&1/36\\
 &10001110&61/9&27/4&1/36\\
 &10110001&61/9&27/4&1/36\\
 &10111100&61/9&27/4&1/36\\
 &11110010&61/9&27/4&1/36\\
\\
9&011000101&245/32&574/75&7/2400\\
\\
10&0100110110&77/9&94/11&1/99\\
  &0110000011&77/9&94/11&1/99\\
  &0110001110&77/9&94/11&1/99\\
  &0110110001&77/9&94/11&1/99\\
  &0110111100&77/9&94/11&1/99\\
  &0111110010&77/9&94/11&1/99\\
\\
11&00101101001&3309/350&8417/891&2369/311850\\
\\
12&101000001101 &1481/143 &559/54 &37/7722\\
  &101001000011 &1481/143 &559/54 &37/7722\\
  &101001001110 &1481/143 &559/54 &37/7722\\
  &101001110001 &1481/143 &559/54 &37/7722\\
  &101001111100 &1481/143 &559/54 &37/7722\\
  &101011001001 &1481/143 &559/54 &37/7722\\
\\
13&0110100111010 &2534/225 &13873/1232 &463/277200\\
\end{array}$$

Source code (in need of optimisation) in GP/PARI:

    f(s,b,i0,k) = { if(#s==1, s[1],
        if(b%2, sum(i=i0,#s,f(setminus(s,[s[i]]), b\2, i, k+1))/#s*(k+1),
                f(vecextract(s,"2.."), b\2, 1, 0)
        ))
    };
    p(b) = { if(b>1, print1(b%2); p(b\2)) };
    g(n) = { d0 = n+1; s = vector(n+1,i,i-1);
        for(b = 0, 2^(n-1)-1,
            v0 = f(s,b,1,0); v1 = f(s, 2^n-1-b, 1, 0);
            if(d0 >= abs(v0-v1), d0 = abs(v0-v1); b0 = b;
                p(b0+2^n); print(" ",v0,"  ",v1,"  ",d0)
        ))
    };
g(13);