As you mention, this is related to the Hardy-Littlewood k-tuple conjecture. In particular, if their conjecture is true, then the primes are not translation-finite. Indeed, it is possible to find an increasing sequence n1 < n2 < n3 < ⋯ so that for every k, the first k nis form an admissible k-tuple. (For example, I think ni = (i+1)! works.) Then, by the k-tuple conjecture, infinitely many such prime constellations exist and so for all k, (X-n1) ∩ (X-n2) ∩ ⋯ ∩ (X-nk) is infinite. (Here and below, X is the set of primes.)
However, maybe we can prove that the primes are not translation finite by some other means. Unfortunately, the technology is not quite good enough to do that. Proving that the primes are not translation finite would, in particular, prove that there exist n1 < n2 such that (X-n1) ∩ (X-n2) is infinite. In particular, this implies that the gap n2-n1 occurs infinitely often in primes, and so pn+1-pn is constant infinitely often. (The standard notation pn indicates the nth prime.)
The best known upper bound for the size of small gaps in primes is that lim infn→∞ (pn+1-pn)/log pn = 0. This was established by Goldston and Yildirim around 2003 and the proof was later simplified. To the best of my knowledge, the best conditional result is by the same authors; they show that given the Elliott-Halberstam conjecture, the prime gap is infinitely often at most 20 or so.
Edit (20/11/2013) : Yesterday James Maynard posted the paper Small gaps between primes on the arxiv in which he shows that for any $m$ there exists a constant $C_m$ such that $$ p_{n+m}-p_n\leq C_m$$ infinitely often. More about this result can be found on Terence Tao's blog, or in this expository article by Andrew Granville.
In Goldston, Pintz, and Yildirim paper Primes in tuples I, they show that under the assumption of the Elliott Halberstam Conjecture,
$$\liminf_{n\rightarrow\infty}p_{n+1}-p_n \leq 16$$
and they leave the following question on page 3:
Question 3. Assuming the Elliott-Halberstam conjecture, can it be proved that
there are three or more primes in admissible k-tuples with large enough k? Even
under the strongest assumptions, our method fails to prove anything about more
than two primes in a given tuple.
From what I understand, the issue is increasing a coefficient from $1$ to $2$.
Let $\mathcal{H}=\left\{ 1,\dots,h_{k}\right\}$ be our admissible set, and suppose that $\max_{i}h_{i}\leq x.$ The approach is to look at the sum
$$\sum_{x<n\leq 2x}\left(\sum_{i=1}^{k}\vartheta\left(n+h_{i}\right)-\log(3x)\right)W(n),$$
where $\vartheta(n)=1_{\mathcal{P}}(n)\log n$, $1_{\mathcal{P}}(n)$ is the indicator function for the primes, and $W(n)$ is a positive weight function. If this sum is positive, then one of the terms must be positive, so for some $x<n\leq2x$ we have
$$\sum_{i=1}^{k}\vartheta\left(n+h_{i}\right)>\log(3x),$$
and since $\log(n+h_{i})\leq\log(3x)$ for all $n$ in our range, it follows that there are at least two indices $i\neq j$ such that
$$\vartheta(n+h_{i}),\ \vartheta(n+h_{j})\neq0.$$
Selberg advocated that in general for ease of calculation one should take a positive weight function to be a square, $W(n)=\lambda(n)^{2},$ so the goal is to prove the inequality
$$\sum_{x<n\leq2x}\sum_{i=1}^{k}\vartheta\left(n+h_{i}\right)\lambda(n)^{2}>\log(3x)\sum_{x<n\leq2x}\lambda(n)^{2}$$
for some choice of $\lambda(n).$ In Goldston, Pintz, and Yildirim's paper, they choose
$$\lambda(n)=\frac{1}{\left(k+l\right)!}\sum_{\begin{array}{c}
d|P(n)\\
d\leq R
\end{array}}\mu(d)\log\left(\frac{R}{d}\right)^{k+l}$$
where $P(n)=\prod_{j=1}^{k}\left(n+h_{j}\right)$, and $R$ depends on $x$. To use the same approach for $3$ terms, we would need to examine the sum
$$\sum_{x<n\leq2x}\left(\sum_{i=1}^{k}\vartheta\left(n+h_{i}\right)-2\log(3x)\right)\lambda(n)^{2},$$
and show that
$$\sum_{x<n\leq2x}\sum_{i=1}^{k}\vartheta\left(n+h_{i}\right)\lambda(n)^{2}>2\log(3x)\sum_{x<n\leq2x}\lambda(n)^{2},$$
for a suitable choice of $\lambda(n)$. Increasing the coefficient to a $2$ seems to be a fundamental issue, and hopefully an expert can explain why this is the case.
Best Answer
In computational complexity theory, most conjectures that two complexity classes are equal (or not equal, as the case may be) can be relativized to an oracle. Sometimes, as in the case of P = NP, one can obtain contradictory relativizations; i.e., there exists an oracle A such that PA = NPA and an oracle B such that PB ≠ NPB.
In the case of contradictory relativizations, it is tempting to hypothesize that if, for example, PB ≠ NPB for "most" oracles B, then P ≠ NP in the "real" (unrelativized) world. This heuristic was seriously proposed by Bennett and Gill as the "random oracle hypothesis," for a specific precise definition of "most oracles." However, the random oracle hypothesis was disproved by Kurtz. Later, another conjecture was proposed along similar lines: the "generic oracle hypothesis," with a different precise definition of "most oracles." But the generic oracle hypothesis was also disproved, by Foster.