Let the geometric progression be $a_1, a_2, a_3, \dots$ with first term $a_1=\frac{1}{8}$ and common ratio $r$.

So, $a_n=a_1 r^{n-1}=\frac{1}{8}r^{n-1}.$

Given: each term is the arithmetic mean of the next two terms.

That means for every term $a_n$, $a_n=\frac{a_{n+1}+a_{n+2}}{2}.$

Substitute GP terms: $\frac{1}{8}r^{n-1}=\frac{\frac{1}{8}r^n+\frac{1}{8}r^{n+1}}{2}.$

Multiply by $16$: $2r^{n-1}=r^n+r^{n+1}.$

Divide by $r^{n-1}$ (valid since $a_2\ne a_1$, so $r\ne 1$, and also GP terms are defined): $2=r+r^2.$

Rearranging, $r^2+r-2=0.$

Factor: $(r-1)(r+2)=0.$

Since $a_2\ne a_1$, we have $r\ne 1$. Hence, $r=-2.$

Now compute the required quantity: $S_{20}-S_{18}=a_{19}+a_{20}.$

Find $a_{19}$ and $a_{20}$: $a_{19}=\frac{1}{8}(-2)^{18}=\frac{2^{18}}{2^3}=2^{15},$ since $(-2)^{18}=2^{18}$.

$a_{20}=\frac{1}{8}(-2)^{19}=\frac{-2^{19}}{2^3}=-2^{16}.$

Therefore, $S_{20}-S_{18}=a_{19}+a_{20}=2^{15}-2^{16}=-2^{15}.$

Comparing with the options: $-2^{15}$ corresponds to Option A.