Number of combinations with given outcomes for a set number of dice rollsHow to find...

One way to address your problem is to use the Inclusion-Exclusion Principle, which will allow us to avoid doing case work. Let's look at your example.
Since there are 36 possible outcomes for each roll of a pair of six-sided dice, there are ${36}^5$ possible outcomes for five rolls of a pair of dice. From these, we must exclude those outcomes in which a sum of 3, a sum of 4, or a sum of 8 is missing. We want to find $|A\cap B\cap C|={36}^5-|A^{\prime }\cup B^{\prime }\cup C^{\prime }|$
Step 2
By the Inclusion-Exclusion Principle,
$|A^{\prime }\cup B^{\prime }\cup C^{\prime }|=|A^{\prime }|+|B^{\prime }|+|C^{\prime }|-|A^{\prime }\cap B^{\prime }|-|A^{\prime }\cap C^{\prime }|-|B^{\prime }\cap C^{\prime }|+|A^{\prime }\cap B^{\prime }\cap C^{\prime }|$
You observed that there are two ways to obtain a sum of 3, three ways to obtain a sum of 4, and five ways to obtain a sum of 8.
|A′|: Since there are two ways to obtain a sum of 3, there are $36-2=34$ ways to not obtain a sum of 3 on each of the five rolls. Hence, $|A^{\prime }|={34}^5$.
|B′|: Since there are three ways to obtain a sum of 4, there are $36-3=33$ ways to not obtain a sum of 4 on each of the five rolls. Hence, $|B^{\prime }|={33}^5$.
|C′|: Since there are five ways to obtain a sum of 8, there are $36-5=31$ ways to not obtain a sum of 8 on each of the five rolls. Hence, $|B^{\prime }|={31}^5$.
$|A^{\prime }\cap C^{\prime }|$ Excluding sums of 3 and 8 eliminates $2+5=7$ possible outcomes for each of the five rolls. Hence, $|A^{\prime }\cap C^{\prime }|=(36-7{)}^5={29}^5$.
$|B^{\prime }\cap C^{\prime }|$: Excluding sums of 4 and 8 eliminates $3+5=8$ possible outcomes for each of the five rolls. Hence, $|A^{\prime }\cap B^{\prime }|=(36-8{)}^5={28}^5$.
$|A^{\prime }\cap B^{\prime }\cap C^{\prime }|$: Excluding sums of 3, 4, and 8 eliminates $2+3+5=10$ possible outcomes for each of the five rolls. Hence, $|A^{\prime }\cap B^{\prime }\cap C^{\prime }|=(36-10{)}^5={26}^5$.
By the Inclusion-Exclusion Principle, the number of outcomes in which sums of 3, 4, and 8 all appear when five dice are rolled is ${36}^5-{34}^5-{33}^5-{31}^5+{31}^5+{29}^5+{28}^5-{26}^5$.