see the diagram i have in post #2 here

from that diagram, you can find the probability of getting a sum of 7 on any one roll. take that as your probability of success, call it p. let q = 1 - p be the probability of failure. then your desired probability, given by the Binomial distribution would be (if we let X be the number of 7's we get):

here you want k = 0

you could also use logic and go around the binomial distribution (though you would be using it implicitly), by finding the probability of not getting a 7, again by the diagram, and cubing the probability, since we want the intersection of the event that we get no 7 with itself, three times