解:$(1)$因为$a,b$互为相反数,所以$a+b=0;$$c $的倒数是$4,$
所以$c=\frac {1}{4}。$则$3a+3b-4c=3(a+b)-4c=3×0-4×(\frac {1}{4})=-1。$
$ (2)$因为$d$的绝对值是最小的正整数,
所以$d=±1。$当$d=1$时,
$8c-d+cd=8×(\frac {1}{4})-1+(\frac {1}{4})×1=2-1+\frac {1}{4}=\frac {5}{4};$
当$d=-1$时,$8c-d+cd=8×(\frac {1}{4})-(-1)+(\frac {1}{4})×(-1)=2+1-\frac {1}{4}=\frac {11}{4}。$
综上,
$ (1)$的值为$-1,$
$ (2)$的值为$\frac {5}{4}$或$\frac {11}{4}。$