jmc

Convert $1d41_8$ to base 10 to get $1d41_8=8^3+8^{2}d+8^1\cdot 4+8^0=512+64d+32+1=545+64d$. Since the possible values of $d$ are 0, 1, 2,..., 7, the possible values of $n$ form an arithmetic series with first term 545 and last term $545+64\cdot 7=993$. The sum of an arithmetic series is $(\text{first term}+\text{last term})(\text{number of terms})/2$, so the sum of the possible values of $n$ is $(545+993)(8)/2=\boxed{6152}$.