Proper PAC learning of 2-DNF under uniform distribution
I don't know if you'll consider the following a non-trivial bound, but here I go.

First, to be clear so that we're not confusing $c$-DNF with $k$-term DNF (which I often do), an $c$-DNF formula over variables $x_1, \ldots, x_n$ is of the form $\vee_{i=1}^{k}(\ell_{i,1} \wedge \ell_{i,2} ... \ell_{i,c})$ where $\forall 1 \le i \le k$ and $1 \le j \le c$, $\ell_{i,j} \in \{x_1, \ldots, x_n, \bar{x}_1, \ldots, \bar{x}_n \}$.

We can first ask how many distinct terms can exist in an $c$-DNF.  Each term will have $c$ of the $n$ variables, each either negated or not -- making for $2^c\binom{n}{c}$ different possible terms.  In a 2-DNF instance, each term will either appear or not, making for $|\mathcal{H}| = 2^{2^c\binom{n}{c}}$ possible "targets," where $\mathcal{H}$ is the hypothesis space.

Imagine an algorithm that takes $m$ samples and then tries all of the $|\mathcal{H}|$ hypotheses until it finds one that predicts perfectly on the samples.  Occam's Razor theorem says that you only need to take about $m = O(\frac{1}{\epsilon}|(\mathcal{H}|+\frac{1}{\delta})$ samples for this algorithm to find a target with error $\le \epsilon$ with probability $\ge 1-\delta$.

In our case, for $c=2$, $\lg|\mathcal{H}| = O(n^2)$, which means you'll need about $n^2$ samples to do the (proper) learning.

But the whole game in learning is not really sample complexity (though that's part of the game, especially in attribute-efficient learning), but rather in trying to design polynomial-time algorithms.  If you don't care about efficiency, then $n^2$ is the simplest answer for PAC sample complexity.

UPDATE (given the changed question):

Because you explicitly stated that you only cared about sample complexity, I presented the brute-force Occam Algorithm, which is the probably the simplest argument.  However, my answer was a bit coy.  $2$-DNF are actually learnable in polynomial time!  This is a result from Valiant's original paper, "A Theory of the Learnable."  In fact $c$-DNF are learnable for any $c = O(1)$.

The argument goes as follows.  You can view a $c$-DNF as a disjunction of $\approx n^c$ 
"meta-variables" and try to learn the disjunction by eliminating the meta-variables inconsistent with the examples.  Such a solution can be easily translated back to a "proper" solution, and takes $O(n^c)$ time.  As a side-note, it is still open whether there is polynomial-time algorithm for $c = \omega(1)$.

As to whether the $n^2$ sample complexity is also a lower bound, the answer is pretty much yes.  This paper by Ehrenfeucht et al. shows that the Occam bound is almost tight.