Solution:
Answer: $800$

Since $360=2^{3} \cdot 3^{2} \cdot 5$, the only possible prime divisors of $d_{i}$ are $2$, $3$, and $5$, so we can write $d_{i}=2^{a_{i}} \cdot 3^{b_{i}} \cdot 5^{c_{i}}$, for nonnegative integers $a_{i}, b_{i}$, and $c_{i}$. Then, $d_{1} d_{2} d_{3} \mid 360$ if and only if the following three inequalities hold.
$$\begin{aligned} a_{1}+a_{2}+a_{3} & \leq 3 \\ b_{1}+b_{2}+b_{3} & \leq 2 \\ c_{1}+c_{2}+c_{3} & \leq 1 \end{aligned}$$
Now, one can count that there are $20$ assignments of $a_{i}$ that satisfy the first inequality, $10$ assignments of $b_{i}$ that satisfy the second inequality, and $4$ assignments of $c_{i}$ that satisfy the third inequality, for a total of $800$ ordered triples $\left(d_{1}, d_{2}, d_{3}\right)$.

(Alternatively, instead of counting, it is possible to show that the number of nonnegative-integer triples $\left(a_{1}, a_{2}, a_{3}\right)$ satisfying $a_{1}+a_{2}+a_{3} \leq n$ equals $\binom{n+3}{3}$, since this is equal to the number of nonnegative-integer quadruplets $(a_{1}, a_{2}, a_{3}, a_{4})$ satisfying $a_{1}+a_{2}+a_{3}+a_{4}=n$.)