The dice has 6 sides.
- Let n be the number of dice rolled at once (8 in the example);
- Let v be the limit at which an individual roll is considered successful (5 or more in this example);
- Let m be the number of successful rolls (3 in our example) that needs to be reached.
The probability of obtaining at least 1 successful roll is: 1-(1-(6-v+1)/6)**n
What would the formula be for the probability of getting at least m successful rolls?
The probability of exactly $m$ "successful" rolls is (from the binomial distribution) $${n \choose m} \dfrac{(6-v+1)^m (v-1)^{n-m}}{6^n}$$ so the probability of at least $m$ "successful" rolls is $$\sum_{j=m}^n {n \choose j} \dfrac{(7-v)^j (v-1)^{n-j}}{6^n}$$ or equivalently $$1-\sum_{j=0}^{m-1} {n \choose j} \dfrac{(7-v)^j (v-1)^{n-j}}{6^n}$$ which is your result when $m=1$
For large $n$ you could use a normal approximation to the binomial distribution with a continuity correction and find an approximation of $$\Phi\left( \dfrac{7n +3 - 6m-nv}{\sqrt{n(7-v)(v-1)}} \right)$$