In this answer I presume that all digits $1,2,5,6$ must be present in the number.
There are $4!=24$ such numbers.
We can index them with $i=1,\dots,24$
and write each of them as $$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$$
where $\left\{ a_{i},b_{i},c_{i},d_{i}\right\}=\{1,2,5,6\} $
Then $$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$$
$\left\{ i\mid a_{i}=1\right\} $, $\left\{ i\mid a_{i}=2\right\} $,
$\left\{ i\mid a_{i}=5\right\} $ and $\left\{ i\mid a_{i}=6\right\} $
have equal cardinality $\frac{24}4=6$ so: $$\sum_{i=1}^{24}a_{i}=6.1+6.2+6.5+6.6=6(1+2+5+6)=84$$
This can also be applied for $b,c$ and $d$ and finally we find:
$$\sum_{i=1}^{24}n_{i}=84+10.84+100.84+1000.84=1111.84=93324$$
Does this really answers your question? If not, then please let me know.
One way is to use Inclusion/Exclusion. You attempted a form of it, but something more elaborate is needed.
You know that there are $\frac{8!}{2!2!2!}$ possible sequences if we have no restriction.
Now let us count the number that have the two $1$'s together. Tie them together, and think of them as a superletter. Now we have $7$ "letters." These can be arranged in $\frac{7!}{2!2!}$ ways. We get the same count for the $2$'s together, and for the $3$'s together. So our first estimate for the required number is $\frac{8!}{2!2!2!}-3\cdot \frac{7!}{2!2!}$.
But we have subtracted too much, for we have subtracted one too many times the patterns that have two $1$'s and two $2$'s together, as well as the ones in which we have two $2$'s and two $3$'s together, and so on. There are $\frac{6!}{2!}$ of each of these, so our next estimate is
$\frac{8!}{2!2!2!}-3\cdot \frac{7!}{2!2!}+3\cdot \frac{6!}{2!}$.
But we have added back too much, for we have added back one too many times the $5!$ sequences in which the $1$'s are together, and the $2$'s, and the $3$'s. Our final count is $\frac{8!}{2!2!2!}-3\cdot \frac{7!}{2!2!}+3\cdot \frac{6!}{2!}-5!$.
Best Answer
If you fix 8 as the last digit, you see that there are $4 \cdot 3 \cdot 2$ ways to complete the number. Thus, 8 appears 24 times as the last digit. By the same logic, if we enumerate all possible numbers using these 5 digits, each number appears 24 times in each of the 4 positions. That is, the digit 8 contributes $(24 \cdot 8 + 240 \cdot 8 + 2400 \cdot 8 + 24000 \cdot 8)$. In total, we have $$(0 + 2 + 3 + 5 + 8)(24 + 240 + 2400 + 24000) = 479952$$ as our total sum.
Update: In case 4-digit numbers cannot start with 0, then we have overcounted. Now we have to subtract the amount by which we overcounted, which is found by answering: "What is the sum of all 3-digit numbers formed by using digits 2,3,5, and 8?" Now if 8 appears as the last digit, then there are 6 ways to complete the number, so 8 contributes $(6\cdot 8 + 60 \cdot 8 + 600 \cdot 8)$. In total, we have $$(2 + 3 + 5 + 8)(6 + 60 + 600) = 11988.$$ Subtracting this from the above gives us 467964.