Using principle of inclusion and exclusion, I've come this far:
Total should be $\binom{8}{2,2,2,2} = 2520$ and now we have to subtract where we counted the intersections together.
so let $S_0 = 5040$
$S_1 = 4* \binom{6}{2} * 4 = 420$
$S_2 = 4* \binom{5}{2} * 3 = 120$
I'm not sure how to calculate $S_3$ which should be the intersection of 1,2,3,and 4
So the answer would be I believe along the lines of 2520 - 420 + 120 -???? + ????
To determine the number of arrangements that contain at least one pair of consecutive equal digits, we can start by considering this pair as a unique item, so we have to rearrange $7$ elements. Since the fixed pair can be chosen in $\binom {4}{1}=4 \,\,$ ways (one for each different digit), we gets our first estimate
$$A_1=4 \cdot \frac {7!}{2!2!2!}= 2520$$
Clearly in this number some arrangements have been counted more than one time because of the intersections. We have then to determine the numbers $A_2$, $A_3$, $A_4$ of arrangements that contain at least $2$, $3$, or $4$ pairs of consecutive equal digits, and then calculate $A_1-A_2+A_3-A_4 \,\,\,\,$ according to the principle of inclusion and exclusion. To determine the number $A_2$ of arrangements that contain at least two pairs of consecutive equal digits, we can again note that these pairs can be considered as two unique items, so we have to rearrange $6$ elements. Since these pairs can be chosen in $\binom {4}{2}=6 \,\,$ ways, we have
$$A_2=6 \cdot \frac {6!}{2!2!}= 1080$$
In a similar way, we obtain
$$A_3=\binom {4}{3} \cdot \frac {5!}{2!}= 240$$ $$A_4=\binom {4}{4} \frac {4!}{1!}= 24$$
So we finally get a total of
$$A_1-A_2+A_3-A_4 \\ =2520-1080+240-24 \\ =1656$$
arrangements that contain two consecutive digits. To get the number of arrangements that contain no two consecutive numbers, subtract this from the total number of possible arrangements that you correctly have calculated and reported in the OP.