It is Possible!

we know that $N \equiv f(N) \equiv f(f(N)) \equiv f(f(f(N))) \pmod{9}$ .

From another side :

$N=1999^{1999} < 2000^{2000} = 2^{2000} \times 10^{2000} =1024^{200} \times 10^{6000}$

$\implies N < 10^{800} \times 10^{6000} = 10^{6800}$

So $N < 10^{6800} \implies f(N) < 6800 \times 9=61200 \implies f(f(N)) < 5 \times 9=45$ $\implies f(f(f(N))) < 2 \times 9 = 18$

we know that $N \equiv f(f(f(N))) \equiv 1999^{1999} \equiv 1^{1999} \equiv 1 \pmod{9}$

So $f(f(f(N))) \equiv 1 \pmod{9}$ and $f(f(f(N))) < 18$

$\implies f(f(f(N)))=10$

so $f(f(f(f(N))))$ is $1$