I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.
Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series. In other words, a trigonometric series is just any series of the form
$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$
This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere. A fourier series is the trigonometric series associated to an
$L^1$ function by taking $A_n$ and $B_n$ as above.
Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point. Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.
A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$. In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.
I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?
P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.
Addendum: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.