I start by knowing that the flip operator is $F=-\sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$\sigma_1^2 = \sigma_2^2=\sigma_3^2=-i\sigma_1\sigma_2\sigma_3 = 1$$
lets us construct $(\sigma_1\sigma_2\sigma_3)^2 = -1$ to get
$$F=-\sigma_1(\sigma_1\sigma_2\sigma_3)^2$$ $$\require{cancel} F=-\cancel{\sigma_1\sigma_1}\sigma_2\sigma_3\sigma_1\sigma_2\sigma_3$$
We can also recover the solution provided by @euler-is-alive because $\sigma_1\sigma_2\sigma_3=i$ is just a number and so we can freely rearrange as $\sigma_1\sigma_2\sigma_3\sigma_2\sigma_3$