You are right that the total number of arrangements is $8!$.
I don't understand your "$8!/3P3$" to "handle" the case of all three vowels appearing consecutively: if all three vowels appear together, then there are six locations for the block of three vowels to be placed at. There are $3!$ ways of placing the vowels in the block, and $5!$ ways of placing the remaining letters in the remaining spots. That gives $6\times 3!\times 5!$ ways in which the three vowels may appear together. But this is neither equal to $8!/3P3$, nor is it equal to $8!-8!/3P3$. So I don't understand what you mean by "handling" the case, or how the case relates to $8!/3P3$.
In how many ways can at least two vowels appear together? Select the location of the block of two vowels, which can be done in $7$ different ways. Then select the vowels to appear in the block, in order: $3\times 2$. Then place the remaining six letters in the remaining six spots: $6!$. We get $7\times 6\times 6!$. But this overcounts: if the three vowels appear together, then we counted that particular arrangement twice: once when we selected the first two vowels as the block, once when we selected the last two vowels as the block.
So: the total number in which no two vowels appear together would be: $(\text{total number of arrangements}) - (\text{arrangements with at least two vowels together}) + (\text{arrangements will all three together}).$