Primes of that form are very rare for n > 2 so you can't expect that they fill an arithmetic progression. Heath-Brown proved that there are infinitely many primes which are sums of two cubes. The proof is hard.
See corrections below...
EDIT, Will Jagy. Evidently Felipe posted from an ipod and was not able to edit at length. He will be able to fix everything when we he settles down his weary feet from his current travels. He meant the bit about such primes being rare. From his comment, it seems most likely that he meant to say originally that there are infinitely many primes of shape $$ x^3 + 2 y^3.$$
Moving on to Victor's comment, primes of shape $ a^3 + b^3$ actually solve
$ p = 3 a^2 - 3 a + 1.$ This means that the number of such primes up to some large positive $M$ is at most $$ c_4 sqrt M$$ which is asymptotically much less than the number of primes up to the same $M$ in some arithmetic progression, that being $$ c_5 frac{M}{log M} $$
So, infinitely many primes, indeed a set of full density, fail to have such an expression.
Finally, it would be a remarkable thing if it were known that there were infinitely many such primes. We certainly do not know the status for primes $n^2 + 1.$ Indeed, I am moved to ask: is there any polynomial in one variable, integer coefficients, and degree at least two that is known to represent infinitely many primes?
He then gives a link to the Friedlander-Iwaniec theorem, maybe I have it right.
http://en.wikipedia.org/wiki/Friedlander%E2%80%93Iwaniec_theorem
For Felipe, Whenever We May Find Him: $$ $$
http://www.youtube.com/watch?v=buMH_tAu2Zg
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