Answer     $\frac{47}{256}$

Let's work out this by case-work. The number of people standing is at most $4$.

• $0$ people standing: $1$ arrangement.
• $1$ people standing: $8$ arrangements
• $2$ people standing: $20$ arrangements (see below)
• $3$ people standing: $16$ arrangements (see below)
• $4$ people standing: $2$ arrangements

Therefore, totally $1+8+20+16+2 = 47$ possibilities exists. Hence the answer is $\frac{47}{2^8} = \boxed{\frac{47}{256}}$.

$2$ people standing

There are $8$ ways to select the first standing person and $5$ ways to select another one who is not adjacent to the first one. However, these two standing people can be chosen in the reverse order, i.e. the duplication count is $2$. Hence, the result is $8\times 5 \div 2 = 20$.

$3$ people standing

This equivalent to count positive integer solutions to $x_1 + x_2+x_3=5$ where $x_i$ represents the number of people sitting between two standing people. The result is $6$. It needs to be multiple by $8$ and then divided by $3$, which leads to $16$. This is because the $1{st}$ person has $8$ choices to pick up his seat. But there are three ways to determine who choose first (due to circular symmetry).