(Corrected and expanded, again!)

As mentioned in the comments, the number of third permutations depends on the relationship between the first two. Asymptotically, the number of third permutations is $\sim e^{-2} n!\, (1-1/n - 1/(2n^2)+O(n^{-3}))$ regardless of the first two permutations. For the exact number, follow Ira's hint: Riordan, *Introduction to Combinatorial Analysis*, chapter 8, part 3.

I can give a hint of how much the number of third permutations varies according to the first two. Let $s$ be the number of intercalates in the first two permutations. (An intercalate is two positions where the two permutations have the same two entries in opposite order: where one has $ab$ the other has $ba$.) Since intercalates cannot overlap, the number of them can't exceed $n/2$. Asymptotically, the number of third permutations which are a derangement of the first two is
$$ e^{-2} n!\, \Bigl( 1 - \frac 1n - \frac 1{2n^2} + \frac 1{3n^3} + \frac {s}{n^4} + O(n^{-4})\Bigr).$$

This is from C. D. Godsil and B. D. McKay, Asymptotic enumeration of Latin rectangles, J. Combinatorial Theory, Ser. B, 48 (1990) 19-44. Corrected version.

A triple of permutations, each two of which are derangements of each other, is a 3-row Latin rectangle. There is a simple summation for the number of them due to Yamamoto, see page 18 in Stone's survey.

Combinatorial Analysisbut I don't have access to my copy right now. $\endgroup$ – Ira Gessel Sep 12 at 21:13