Lower bound on the support size of an $\epsilon$-biased distribution Let $D$ be an $\epsilon$-biased distribution we want to show that
$$\text{Supp}(D)\geq \Omega\bigg(\frac{n}{\epsilon^2\log(\frac{1}{\epsilon})}\bigg)$$
I know that there are some proofs for this but I am intrested in the following particular way.

Let $X,Y$ be two distributions over $\{0,1\}^n$. Consider the "convolution" of distributions $Z=X\circledast Y$ to be defined by sampling $x\sim X$ and $y\sim Y$ independently and outputting $x\oplus y$ where $\oplus$ is just bit-wise xor.

Now if we define a distribution $D^{(t)}=D\circledast\cdots\circledast D$ ($t$ times) we can use some Fourier analysis and combinatorics to show that


$D^{(t)}$ is $\epsilon^t$-biased.
$|\text{Supp}(D^{(t)})|\leq\binom{|\text{Supp}(D)|+t}{t}$


We can also show that for every $\delta$-biased distribution $Z$ we have $\lVert Z\rVert_2^2\leq\delta^2+2^{-n}$ when considering the distribution as a vector of probabilities. Now, I was told that I can use the latter to obtain a lower bound on $|\text{Supp}(D^{(t)})|$ and then choose $t$ accordingly to get the desired bound.

However, the only bound I can get for $|\text{Supp}(D^{(t)})|$ is 
$$|\text{Supp}(D^{(t)})|\geq\sqrt{\frac{1}{\epsilon^{2t}+2^{-n}}}$$
And I don't see how that helps...
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.