This is just a partial answer in that it gives the probability mass function for finite values of $n$ (i.e., no asymptotics).
Using Mathematica one can find the probability mass function for a specific $n$ and $\beta$ with the following commands:
distY = OrderDistribution[{BinomialDistribution[n, 1/2], 2^(beta n)}, 1];
pmf = PDF[distY, y]
with the following result:
$$Pr(Y=y)=\left(2^{-n} \binom{n}{y}-I_{\frac{1}{2}}(n-y,y+1)+1\right){}^{2^{\beta n}}-\left(1-I_{\frac{1}{2}}(n-y,y+1)\right){}^{2^{\beta n}}$$
when $0\leq y<n$ and
$$Pr(Y=n)=\left(2^{-n} \binom{n}{y}\right)^{2^{\beta n}}$$
when $y=n$ where $I_z (a,b)$ is the regularized incomplete beta function.
If $\beta=1/20$, then one can construct a table of the means and variances for values of $n$:
beta = 1/20;
distY = OrderDistribution[{BinomialDistribution[n, 1/2], 2^(beta n)}, 1];
data = Table[{n, 2^(beta n), Mean[distY] // N, Variance[distY] // N}, {n, 20, 240, 20}]
TableForm[data, TableHeadings -> {None, {"n", "k", "Mean", "Variance"}}]
$$
\begin{array}{cccc}
20 & 2 & 8.74629 & 3.42822 \\
40 & 4 & 16.7558 & 4.91162 \\
60 & 8 & 24.5034 & 5.56277 \\
80 & 16 & 32.1274 & 5.85087 \\
100 & 32 & 39.6867 & 5.97801 \\
120 & 64 & 47.2088 & 6.03288 \\
140 & 128 & 54.7079 & 6.05502 \\
160 & 256 & 62.1913 & 6.06233 \\
180 & 512 & 69.6636 & 6.06297 \\
200 & 1024 & 77.1272 & 6.06068 \\
220 & 2048 & 84.5839 & 6.05716 \\
240 & 4096 & 92.0349 & 6.05319 \\
\end{array}
$$
With larger values of $n$, there will likely be numerical overflow issues unless some care is taken. However, so far the relationship with $n$ and the mean seems pretty linear:
