You can do something with simple scaling as long as you work on the full space. Assume that $|\cdot|$ and $\|\cdot\|$ are two shift-invariant (semi)norms and for the scaling operator $T_a f(x)=f(x/a)$ you have $|T_af|=a^t|f|$, $\|T_a f\|=a^s\|f\|$. Then, if you do the $\delta$-mollifying on $f$, it is equivalent to the $a\delta$-mollifying on $T_a f$, which means that if any inequality of the form $|f-f_\delta|\le C\delta^r\|f\|$ exists at all, you must have $a^t=a^r a^s$, i.e., $r=t-s$.
Now, to check that the inequality is there, you just need to check that $\|f\|$ dominates the deviation of $f$ from some faithfully reproduced function in the norm $|\cdot|$. Note that we need the homogeneous spaces for such scaling tricks, so the Sobolev norm will really mean the $L^p$ norm of the highest derivative involved, not the full sum of $L^p$-norms of the previous derivatives.
Example: $C$ and $C^k$ (with uniform bounds on the entire line). If we have any chance to get anything at all, the speed is $\delta^k$ by scaling. We have this chance realized if and only if our mollifier (which I assume to be compactly supported for simplicity) has first $k-1$ moments correct (i.e., reproduces the polynomials of degree up to $k-1$ precisely).
