UPDATE 9/14/2018
I think it might still be unknown whether the constant can be reduced below $e$
With Alexandros Eskenazis we came up with an argument which improves the bound $\|f\|_{2}\leq e^{k}\|f\|_{1}$ to $\|f\|_{2}\leq (2.69075...)^{k}\|f\|_{1}$ for polynomials $f$ of degree $k$ on the Hamming cube. I suspect one can further improve the bound but then one needs to carefully play with conformal maps. Anyways, here is the proof which in a sense corresponds to "averaging" the previous argument.
Proof. Take any $q>2$ and consider the following domain in the complex plane
$$
\Omega := \left\{ z \in \mathbb{C}\, :\, \left|z\pm \frac{q-2}{2(q-1)}\right|\leq \frac{q}{2(q-1)}\right\}.
$$
Let $p$ be the conjugate exponent to $q$. It follows from complex hypercontractivity that for any $f:\{-1,1\}^{n} \to \mathbb{C}$, $f(x) = \sum_{S \in \{1,2,\ldots, n\}} a_{S} W_{S}(x)$, where $W_{S}(x)=\prod_{j \in S}x_{j}$, $x=(x_{1}, \ldots, x_{n})$ are Walsh functions, we have
$$
\|T_{z}f\|_{q} \leq \|f\|_{p}
$$
whenever $z \in \Omega$, where $T_{z}f(x) = \sum_{S \in \{1,2,\ldots, n\}} z^{|S|}a_{S} W_{S}(x)$. Let $\pi \beta$ be the exterior angle (measured in radians) of the domain $\Omega$ at point $z=\frac{i}{q-1}$. Clearly $\beta = \frac{1}{2}+ \frac{1}{\pi}\arctan\left(\frac{q-2}{2}\right)$. Next, consider the conformal map
$$
\varphi(z) = \frac{1+\left(\frac{zi\sqrt{q-1}+1}{zi\sqrt{q-1}-1}\right)^{1/2\beta}}{1-\left(\frac{zi\sqrt{q-1}+1}{zi\sqrt{q-1}-1}\right)^{1/2\beta}}.
$$
Notice that $\varphi$ maps conformally the comploment of $\Omega$ onto the complement of the unit disk in $\mathbb{C}$, and it has a linear growth at infinity. Therefore for any polynomial $p(z)$ of degree $k$ on $\Omega$ we see that the map $z \mapsto \frac{p(z)}{\varphi(z)^{k}\|p\|_{C(\Omega)}}$ is regular at infinity and it is bounded by $1$ on $\partial \Omega$. Therefore by the maximum principle we have $|p(z)|\leq |\varphi(z)|^{k}\|p\|_{C(\Omega)}$ for all $z \in \Omega^{c}$. In particular this means that
$$
|p(1)|\leq |\varphi(1)|^{k}\|p\|_{C(\Omega)} \qquad (***)
$$
Next, let $V$ be the vector space of all polynomials of degree $k$ on $\Omega$. It is the subspace of $C(\Omega)$. Let $L$ be the linear functional on $V$ which acts as follows: for any $p \in V$, $L(p)=p(1)$. It follows from (***) that $|L(p)|\leq |\varphi(1)|^{k}\|p\|_{C(\Omega)}$. By Hahn--Banach theorem $L$ extends to a bounded functional $\tilde{L} \in C^{*}(\Omega)$ so that $\|\tilde{L}\|_{C^{*}(\Omega)}\leq |\varphi(1)|^{k}$, and $\tilde{L}|_{V}=L$. The space $C^{*}(\Omega)$ can be identified to the Banach space of complex Radon measures on $\Omega$ equipped with total vartiation norm, so that $\tilde{L}(g) = \int_{\Omega} g d\mu$ for all $g \in C(\Omega)$, and $|\mu|(\Omega):=\|\mu\|_{TV} = \|\tilde{L}\|_{C^{*}(\Omega)}\leq |\varphi(1)|^{k}$.
Now pick any $f:\{-1,1\}^{n} \to \mathbb{C}$, and consider the polynomial $p(z)=T_{z}f(x)$. We have
$$
\left\|f\right\|_{q} = \left\|\int_{\Omega}T_{z}f d\mu(z) \right\|_{q} \leq \int_{\Omega} \left\|T_{z}f \right\|_{q}d|\mu|(z)\\
\leq \int_{\Omega} \left\|f \right\|_{p}d|\mu|(z) \leq \|\mu\|_{TV} \|f\|_{p}\leq |\varphi(1)|^{k} \|f\|_{p}
$$
for all $q\geq 2$ and the conjugate $p$. Using log-covexity of $L^{p}$ norms we obtain
$$
\|f\|_{2} \leq \|f\|_{1} \left(\frac{\|f\|_{q}}{\|f\|_{p}}\right)^{\frac{1}{2(1/p-1/q)}} = \|f\|_{1} |\varphi(1)|^{\frac{k}{2(1/p-1/q)}}
$$
Now if my calculation is correct we have
$$
|\varphi(1)|^{\frac{k}{2(1/p-1/q)}} = \left(\frac{1+\cos\left(\frac{\arctan\left(\frac{2\sqrt{q-1}}{q-2}\right)}{1+\frac{2}{\pi}\arctan\left(\frac{q-2}{2}\right)}\right)}{1-\cos\left(\frac{\arctan\left(\frac{2\sqrt{q-1}}{q-2}\right)}{1+\frac{2}{\pi}\arctan\left(\frac{q-2}{2}\right)}\right)}\right)^{\frac{k}{4(1/p-1/q)}},
$$
optimizing the latter quantity over all $q>2$, and using the fact that $p=\frac{q}{q-1}$, we see that the minimal value is attained around the point $q=2.39079...$ which gives the value $(2.69075...)^{k}$.
