Define $f(n)=(2n)^2 + 1,n \in \mathbb{N}$
From $1$ to $10^7$ there's $15$ numbers that $f(n)$ is prime, $f(f(n)), f(f(f(n)))$ and $f(f(f(f(n))))$ are also primes.
The $15$ numbers are:
625678, 704613, 717718, 1182168, 3147353,
4869813, 5339178, 5363578, 5411562, 846777,
7848283, 7970403, 8152962, 9220303, 9727978
Is that normal $4$ numbers in this $15$ numbers end in $78$?
On
There are $40$ possible final two digits -those that end $2, 3, 7$ and $8$
The chance that one of the $40$ appears four times is $$40{15\choose4}0.025^40.975^{11}=0.016$$
So it's a bit surprising, but not if you have done half a dozen of these functions. There are also other things that might have drawn your attention if the numbers had come out differently - for example several close together.
To expand on my comment, here are the first 15 values of f(n):
$$5,17,37,65,101,145,197,257,325,401,485,577,677,785,901$$ Because of symmetry for certain primes (those congruent to $1 \bmod 4$), this can test those up as high as 29 (possibly higher) if you do mod these by those primes ($5,13,17,29$) you get the following sequences (respectively):$$0,2,2,0,1,0,2,2,0,1,0,2,2,0,1$$ $$5,4,11,0,10,2,2,10,0,11,4,5,1,5,4$$ $$5,0,3,14,16,9,10,2,2,10,9,16,14,3,0$$ $$5,17,8,7,14,0,23,25,6, 24, 21, 26, 10, 2, 2$$ disallowing: $$1,4\bmod 5$$$$4,9\bmod 13$$$$2,15\bmod 17$$$$6,23\bmod 29$$ as start values( in fact, at best end iterate values if 1 more than a quadratic residue), continuing we get: $$0\bmod 5$$$$2,11\bmod 13$$$$8,9\bmod 17$$$$ 8,10,19,21\bmod 29$$ as illegal prior to the second last iterate : $$1,4\bmod 5$$$$3,6,7,10\bmod 13$$$$6,11\bmod 17$$$$3,11,13,16,18,26\bmod 29$$ for third last iterate. Taking the complement, of the union,for each prime; we get:$$2,3\bmod 5$$$$0,1,5,8,12\bmod 13$$$$0,1,3,4,5,7,10,12,13,14,16\bmod 17$$$$0,1,2,4,5,7,9,12,14,15,17,20,22,24,25,27,28\bmod 29$$ Long story short is just 2 residues mod 5, 5 residues mod 13, 11 residues mod 17, and 17 residues mod 29 survive as possible start residues, for 1870 residues ( fixed calculation error in the edit, aka under 5.9% of residue classes ) mod their product (32045) and 20 times their product is a multiple of 100 so you can go from there.
EDIT
Turns out it's just as likely as any other ending via this, you just seem to be getting roughly 1 every multiple of 640900. might have to do with placement in multiples of 640900 if any of these numbers where divisible by a prime that is of form $4x+3$, we get it divisible by an even number of them. Any ending that is 0 mod 3, shows up only if 640900 isn't multiplied by a multiple of 3, and non-zero residue two digit endings don't happen when 640900 multiplied by their additive inverse mod 3. Might explain some of why they are spread out. maybe some multipliers are just denser in working values, when you consider other factors.