Lower bound on the support size of an $\epsilon$-biased distribution
You shouldn't have a square root. Namely, for every $\delta$-biased distribution $Z$ (using your notation), we have
$$
\delta^2+2^{-n} \geq \lVert Z\rVert^2_2 \geq \frac{1}{\lvert\operatorname{supp} Z\rvert}\tag{1}
$$
since the squared $\ell_2$ norm of a distribution over support of size $N$ is at least that of the uniform distribution on $N$ elements, which is $1/N$.
This gives you, for $t\geq 1$ to choose suitably, 
$$\lvert\operatorname{supp} D^{(t)}\rvert \geq \frac{1}{\varepsilon^{2t}+2^{-n}} \tag{2}
$$
since as you have shown $D^{(t)}$ is $\delta$-biased for $\delta=\varepsilon^t$. A natural choice of $t$, in view of (2), is the one balancing the denominator, i.e., such that $\varepsilon^{2t}=2^{-n}$. This leads to taking
$$
t \stackrel{\rm def}{=} \frac{n}{2\log(1/\varepsilon)}\tag{3}
$$
Recalling the standard bound $\binom{n}{k}\leq \left(\frac{en}{k}\right)^k$ for $1\leq k\leq n$ along with (2), (3), and what you mentioned as 2. in your question, we thus get
$$
\left(\frac{e(\lvert\operatorname{supp} D \rvert+t)}{t}\right)^t \geq \binom{\lvert\operatorname{supp} D \rvert+t}{t} \geq \frac{1}{\varepsilon^{2t}+2^{-n}} = \frac{1}{2\varepsilon^{2t}} \tag{4}
$$
so that
$$
 \frac{e(\lvert\operatorname{supp} D \rvert+t)}{t} \geq \frac{1}{2^{1/t}\varepsilon^{2}} \tag{5}
$$


Rearranging and recalling again the choice of $t$ in (3) yields
$$
 \lvert\operatorname{supp} D \rvert\geq  \Omega\!\left(\frac{n}{\varepsilon^{2}\log(1/\varepsilon)} \right) \tag{6}
$$
as claimed.