Fedja is absolutely right: this has been proven, for sufficiently large $n$, by Drmota, Mauduit and Rivat.
Although it looks at first sight as though this question is as hopeless as any other famous open problem on primes, it is easy to explain why this is not the case. Of the numbers between $1$ and $N := 2^{2n}$, the proportion whose digit sum is precisely $n$ is a constant over $\sqrt{\log N}$. These numbers are therefore quite "dense", and there is a technique in prime number theory called the method of bilinear sums (or the method of Type I/II sums) which in principle allow one to seriously think about finding primes in such a set. This is what Drmota, Mauduit and Rivat do.
I do not believe that their method has currently been pushed as far as (for example) showing that there are infinitely many primes with no 0 when written in base 1000000.
Let me also remark that they depend in a really weird way on some specific properties of these digit representation functions, in particular concerning the sum of the absolute values of their Fourier coefficients, which is surprisingly small. That is, it is not the case that they treat these Hamming sets as though they were "typical" sets of density $1/\sqrt{\log N}$.
I think one might also mention a celebrated paper of Friedlander and Iwaniec, http://arxiv.org/abs/math/9811185. In this work they prove that there are infinitely many primes of the form $x^2 + y^4$. This sequence has density just $c/N^{1/4}$ in the numbers up to $N$, so the analysis necessary to make the bilinear forms method work is really tough. Slightly later, Heath-Brown adapted their ideas to handle $x^3 + 2y^3$. Maybe that's in some sense the sparsest explicit sequence in which infinitely many primes are known (except of course for silly sequences like $s_n$ equals the first prime bigger than $2^{2^n}$).
Finally, let me add the following: proving that, for some fixed $n$, there are infinitely many primes which are the sum of $n$ powers of two - this is almost certainly an open problem of the same kind of difficulty as Mersenne primes and so on.
This question is extremely close to this one
Covering the primes by 3-term APs ?
though not exactly the same.
For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.
To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.
For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.
I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.
Best Answer
The problem may have connections with 3x+1 problem. To an integer, I can associate $(p_0,p_1,\cdots)$ the sequence of the least prime divisor of the term of the sequence $u_{n+1}=f(u_n)$ which may have same properties as the parity function in the 3x+1 problem. For instance : the asymptotic periodicity of $(p_n)$ is equivalent to periodicity of $u_n$ (thus convergence). I would like to prove that for all composite integer,there exists a prime that appear infinitly many times in the sequence $(p_n)$. Using some asymptotic estimations, it's not difficult to prove: $\forall N>0,\exists p | card \lbrace n \in \mathbb N | p_n=p\geq N\rbrace $. It makes no use of the fact that $p_n$ is the least divisor of $u_n$.
It would be interesting to have two other theorems "3x+1"-like : The sequence $(p_n)$ determines $u_0$ and $\sum_{n=0}^{\infty}\prod_{k\leq n} \frac{p_k}{p_k +1}=u_0$ The sum converges since we have the inequality $\sum_{n=0}^{\infty} \prod_{k\leq n} \frac{p_k}{p_k +1}\leq u_0$.
I would be very interested by the links between the choice of a sequence $(p_n)$ and the convergence in an appropriate space of the prime sum. In the "3x+1"-equivalent would be $\sum \frac{2^{a_k}}{3^k}$, where $a_k=$ number of even terms before the $k^{th}$ odd term, which converges in $\mathbb{Z}_2$. In the 3x+1 problem generalized to $\mathbb{Z}_2$, the sum establish a bijection between parity functions and initial conditions in $\mathbb{Z}_2$.
Furthermore is it true that for any k-uplet of prime number $(l_0,\cdots,l_N)$ there exists an initial condition such that $p_i=l_i, i\leq N $.