For all $x>90$, I assert that any $x^2+1$ prime may be written as the sum of five smaller $x^2+1$ primes. In fact, above that bound, I think a stronger conjecture holds where one of said primes can always be $5$.
e.g.
$$\begin{array}\\ 26^2+1&=677\\ &=401+197+37+37+5 \\ &=(20^2+1)+(14^2+1)+2(6^2+1)+(2^2+1) \end{array}$$
The only exceptions appear to be $x\in\{1,2,16,20,90\}$. The primes need not be distinct, although it seems that there is always at least one such representation for $x>170$.
All statements above have been confirmed up to at least $10^6$ with an apparent strong trend toward increasing numbers of representations. Below is a plot of the number of different representations for each $x^2+1$ prime, found by a brute force check using any five $x^2+1$ primes up through $1060^2+1=1123601$ (the $117$th Landau prime). As expected, it only begins tapering off around that bound, and if I hadn't stopped there, I'd expect both minima and maxima to continue to rise. As an informal observation, it also seemed that the largest prime representable was gradually approaching $5(x^2+1)$.
I'm interested in insight on the conjecture and/or confirmation/counterexamples. I am also curious whether this is related to (or even follows from) Lagrange's four-square theorem.
This is just a long comment, regarding the OP's stronger conjecture.
If $5+(A^2+1)+(B^2+1)+(C^2+1)+(D^2+1)=(4n^2+1)$, then
$$A^2+B^2+C^2+D^2=4(n^2-2)$$
Note that if any of the squares is odd, then all four must be odd. Since $1$ is the only odd square that is $1$ less than a prime, and $5+2+2+2+2=13$ is not $1$ more than a square, the variables $A$, $B$, $C$, and $D$ must all be even, so we can rewrite things as
$$a^2+b^2+c^2+d^2=n^2-2$$
where we want $4a^2+1$, $4b^2+1$, $4c^2+1$, and $4d^2+1$ to all be prime.
Now Jacobi's four-squares theorem tells us that the number of integer solutions of $a^2+b^2+c^2+d^2=N$ is
$$r_4(N)=8\sigma(N)-32\sigma(N/4)$$
where $\sigma(N/4)=0$ if $4\not\mid N$. For us, $N=n^2-2$ is never divisible by $4$, so $r_4(n^2-2)=8\sigma(n^2-2)$.
Note that $r_4(N)$ counts solutions with squares of negative numbers as well as squares of positive numbers, and it also counts each arrangement of the squares, so the number of ordered solutions, with $0\lt a\le b\le c\le d$, could be as small as $r_4(n^2-2)/(16\cdot24)={1\over48}\sigma(n^2-2)$.
The easiest way to keep the number of solutions small -- which can be thought of as the easiest way to prevent the four squares from all "accidentally" being $1$ less than a prime -- is for $n^2-2$ to be a prime or twice a prime.
For the OP's reported example, $4n^2+1=2917$, we have $n^2-2=727$, which is a prime. It's also congruent to $7$ mod $8$, so it cannot be written as a sum of three squares. Consequently, the number of solutions in positive integers is $8\cdot728/16=364$. The number of ordered solutions (with $0\lt a\le b\le c\le d$) is somewhere between this and $\lceil364/24.\rceil=16$.
It's maybe worth doing a very crude heuristic argument on the "probability" that $2917$ is not the sum of $5$ plus four primes each $1$ greater than a square, versus the same for $1008017$ (which is the first example past a million). According to OEIS, there are $11$ relevant primes (i.e., disregarding $2$) less than $2917$ and $111$ less than $1008017$.
To the extent that a square participating in a four-square sum for a number $N$ is picked at random from among the squares less than $N$, the probability that all four squares in $4a^2+4b^2+4c^2+4d^2=4\cdot727$ are $1$ less than a prime is $(11/26)^4\approx0.032$. (The denominator is the number of positive squares less than $727$, i.e., $\lfloor\sqrt{727}\rfloor$.) This suggests that the probability that no solution has all four squares of that form is somewhere between $(1-(11/26)^4)^{364}\approx7\cdot10^{-6}$ and $(1-(11/26)^4)^{16}\approx0.5939$.
That lower bound, however, is much too small: any four squares summing to $727$ must have at least two different squares, so the number of ordered solutions is at most $364/4=91$, and thus the lower bound becomes $(1-(11/26)^4)^{91}\approx0.056$, and this suggests that $2917$ has a possibly small, but nonetheless non-negligible shot at not being a sum of $5$ and four other primes that are $1$ more than a square.
For $4n^2+1=1008017$, on the other hand, we have $n^2-2=252002=2\cdot126001$, where $126001$, like $727$, is prime, and thus $r_4(252002)=8\cdot3\cdot126002$. In this case, since there are $501$ positive squares less than $252002$, the corresponding upper bound on the probability is
$$(1-(111/501)^4)^{\lceil126002/16\rceil}=(1-(111/501)^4)^{7876}\approx5.6\cdot10^{-9}$$
and this really is quite small (about seven orders of magnitude smaller than the lower bound for $2917$).
These crude heuristics shouldn't be taken too seriously; they certainly don't prove anything. But they do seem suggestive that primes of the form $x^2+1$ may be sufficiently plentiful that five-fold sums of them (or four-fold sums plus $5$) may indeed cover all sufficiently large primes of the same form -- it's conceivable, in fact, that they cover all sufficiently large numbers of the form $4n^2+1$, prime or not. For that to be the case, of course, we would need an infinite supply of primes of the form $x^2+1$, which is, of course, not known to be the case. The OP's conjecture, on the other hand, does not require an infinite number of such primes. Indeed, it might benefit from the supply running out!