Use the central limit theorem and a continuity correction.
The expected number of $6$s is $360/6=60$. The standard deviation of the number of $6$s is $\sqrt{360\cdot(1/6)\cdot(5/6)}= \sqrt{50} = 5\sqrt{2}$.
"70 or more" is the same as "more than 69". That's where the "continuity correction" tells you to use $69.5$ when you use a continuous approximation to this discrete distribution.
So $\frac{69.5-\text{expected value}}{\text{standard deviation}} = \frac{69.5-60}{5\sqrt{2}} \approx 1.3435.$ You're asking: what's the probability of being more than that many standard deviations above the mean. Plug that number into the appropriate table or software for the normal distribution.
BTW, I'd have said "a fair die", and used "dice" only as the plural.
Later edit: I find it being pointed out that calculators aren't allowed, as if that means this is all of no use. But the answer was phrased as "between" something and something. So you shouldn't need a calculator to apply this; you don't need a precise number, but only that it's between something and something else. You don't need a calculator to know immediately that $\sqrt{50}$ is slightly more than $7$, so that $9.5/\sqrt{50}$ is somewhat more than $1$. Even before doing any of this you know that the probability is less than $0.5$, since $70$ is well above the expected value. And "somewhat more than $1$ is certainly less than $2$. In this case, all you really need is that it's between $1$ and $2$ standard deviations above the mean. So what you see above should do it even if you don't have a calculator.