It is a theorem of Ben Green that every subset of the primes of positive relative density contains a progression of length three. As an immediate consequence, the set of primes $A$ which are not the first term in a progression of primes of length three has density zero (otherwise $A$ would contain a length three progression, a contradiction).

Ben's proof is, strictly speaking, an application of the circle method, but is probably overkill for your problem (roughly speaking, Ben wants to find a length three progression with all three elements in an arbitrary dense subset $A$ of the primes; for your problem, one only needs to study the simpler problem where the smallest term of the progression needs to be in $A$ but the other two elements lie in the set $P$ of primes). So a simpler proof (still by the circle method) is likely to exist. (Note, by the way, that a later paper of Ben and myself gives a slightly simpler proof of Ben's theorem (and, of course, we also have a more complicated proof as well).)

Another approach is to modify the argument of Montgomery and Vaughan (also, ultimately, based on the circle method) that shows that the number of even numbers that are not the sum $p_1+p_2$ of two primes is very low in density (much lower than the density of the primes, in particular). (Actually, an older and somewhat simpler paper of Vaughan already suffices for this.) The same argument should also show that the odd integers $n$ that are not the first term $2p_1-p_2$ of an 3-term AP whose other two terms $p_1,p_2$ are primes larger than $n$, also has density much smaller than that of the primes.