Solution:

Let $n=8$.

First, consider any given student $S$ and an antigen $a$ foreign to him/her. Assuming $S$ has been bitten, we claim the probability $S$ will suffer due to $a$ is
$$1 - \frac{2^{2^{n-1}+1} - 1}{2^{2^{n-1}} (2^{n-1} + 1)}$$
Indeed, let $N = 2^{n-1}$ denote the number of students with $a$. So considering just these students and summing over the number bitten, we obtain a probability
$$\frac{1}{2^{N}} \sum_{t=0}^{N} \binom{N}{t} \binom{N}{t} \frac{t}{t+1} = \frac{1}{2^{N}} \frac{2^{N} N - 2^{N} + 1}{N+1}$$
We now use linearity over all pairs $(S, a)$ of students $S$ and antigens $a$ foreign to them. Noting that each student is bitten with probability $\frac{1}{2}$, and retaining the notation $N=2^{n-1}$, we get
$$\frac{1}{2} \sum_{k=0}^{n} \left[ \binom{n}{k} \cdot k \left( \frac{2^{N} N - 2^{N} + 1}{2^{N}(N+1)} \right) \right] = \frac{n N (2^{N} N - 2^{N} + 1)}{2^{N+1}(N+1)}.$$
Finally, setting $n=8=2^{3}$ and $N=2^{n-1}=2^{7}=128$, we get the claimed answer.