If $p\in\Bbb Z[X]$ show that: $ \max\limits_{x\in [0,1]}\left|p(x) \right| > \frac{1}{e^{n}}. $

Let $ M = \max_{x\in [0,1]}\left|p(x) \right| $ we know that

$$ M = \max_{x\in [0,1]}\left|p(x)\right| =\lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^j dx\right)^{\frac{1}{j}}=\lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}}$$

We then consider $$ M_{j} =\int_0^1 \left|p(x)\right|^{2j} dx$$

since $\deg p= n$ then it follows that , $\deg p^{2j}= 2jn$. so, If we Can write it as, $$p^{2j}(x) = \sum_{k=0}^{2jn} a_{j,k}x^k$$

Therefore, $$ M_{j} =\int_0^1 \left|p(x)\right|^{2j} dx \ge\left|\int_0^1 p(x)^{2j} dx \right| = \left|\sum_{k=0}^{2jn} \frac{a_{j,k}}{i+1}\right|$$ Let $\color{blue}{\operatorname{lcm}(1,2,3,\cdots, 2jn+1)}$ be the least common factor of $(1,2,3,\cdots, 2jn)$. Then since $a_{k,i}\in \Bbb Z$ we have,

$$ M_{j}\ge \left|\sum_{k=0}^{2jn} \frac{a_{j,k}}{i+1}\right|\ge \frac{1}{lcm(1,2,3,\cdots, 2jn+1)}$$ We get from this that, $$\color{red}{\log (\color{blue}{\operatorname{lcm}(1,2,3,\cdots, 2jn+1)})\sim 2jn+1}$$

hence for $\varepsilon>0$ and for $j$ large enough we have,

$$\color{blue}{\operatorname{lcm}(1,2,3,\cdots, 2jn+1)}\le \exp((1+\varepsilon)(2jn+1) ) $$ which implies that

$$ \left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}} = M_j^{1/2j} \ge \left(\frac{1}{\color{blue}{\operatorname{lcm}(1,2,3,\cdots, 2jn+1)}}\right)^{\frac{1}{2j}} \ge \frac{1}{\exp((1+\varepsilon)(n+\frac{1}{2j}) ) }$$

Taking $j\to \infty$ we get

$$ M = \max_{x\in [0,1]}\left|p(x)\right| = \lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}} \ge \frac{1}{\exp((1+\varepsilon)n ) }~~~\forall ~~\varepsilon>0$$

That is $$ \color{blue}{M = \max_{x\in [0,1]}\left|p(x)\right| \ge \frac{1}{e^n }}$$

I have to remark that the given inequality is fairly simple to prove if we replace $e$ with $4$.
Since the shifted Legendre polynomials provide an orthogonal base of $L^2(0,1)$ with respect to the standard inner product, we have $$ \min_{\deg q<n}\int_{0}^{1}\left(x^n+q(x)\right)^2\,dx = \frac{1}{(2n+1)\binom{2n}{n}^2} $$ hence $p(x)\in\mathbb{Z}[x]$ and $\deg p=n$ imply $$ \max_{x\in[0,1]}\left| p(x) \right| \geq \frac{1}{\binom{2n}{n}\sqrt{2n+1}} >\frac{1}{4^n}.$$