Error reduction with expanders and derandomization
You can't expect this to be true since you can reduce the error of a constant-error algorithm to $1/n$ using an additional $O(\log n)$ bits using expanders (and by the way you don't really need even these extra bits to do that). So if an error of $1/n$ can go down to $\exp(-n)$ error using only $O(n/\log n)$ additional bits, then you could always go from constant error to $\exp(-n)$ error using $O(\log n)+O(n/\log n)=O(n/\log n)$.