Norms on $\mathcal{P}_N$ Vector Space of Polynomials up to Order N 
$\|p\|_\infty :=\sup_{x\in [0,1]}|p(x)|$ and $\|p\|_{L^1}:=\int_0^1 |p(x)| dx$.
As the space of real-valued polynomials on $[0,1]$ up to order $N$ is a $N+1$ dimensional vector space and $\|\cdot\|_{\infty}$ and $\|\cdot\|_{L^1}$ are norms for it, there is a constant $C_N$ s.t. $\|\cdot\|_\infty \leq C_N \|\cdot\|_{L^1}$.

Is there an easy way how to compute this $C_N$?
 A: Let us consider the unit sphere under $|| \cdot ||_1$: all polynomials $p(x)$ such that $\int_0^1|p(x)|dx \le 1$.
We have $$\int_0^1 |p(x)| dx \le \int_0^1 \sum |a_i||x^i| dx = \int_0^1 \sum |a_i|x^i dx = |a_0|+\frac{1}{2}|a_1| + \dots + \frac{1}{n+1}|a_n| $$
by the triangle inequality and similar standard inequalities.
Thus, for any polynomial on the L1 unit sphere, we have $$1 \le |a_0| + \frac{1}{2}|a_1| + \dots + \frac{1}{n+1}|a_n|$$
EDIT: Compare the upper bound with the value of the infinity norm: $$||p||_{\infty} = \max |a_0+\dots + a_nx^n| \le \max \sum |a_i x^i| \le \sum |a_i| \le (n+1)\left[|a_0| + \frac{1}{2}|a_i| + \dots \frac{1}{n+1}|a_n| \right]$$
These inequalities give us an upper bound for the infinity norm of any polynomial on the L1 unit ball: namely because any polynomial on the L1 unit ball has to satisfy the first inequality, and we can then use the same input data (i.e. the coefficients) to get an upper bound for the infinity norm of that polynomial.
So ideally we are looking for a polynomial which satisfies the first inequality and which attains the upper bound given by the second inequality -- since it attains an upper bound for the infinity norm of a polynomial on the L1 unit sphere, it must be the maximum, and we are done. 
If we have a polynomial such that $1=\int_0^1 |p(x)| dx = |a_0| + \dots + \frac{1}{n+1}|a_n|$, then it must be on the $L1$ unit sphere. For any other polynomial on the L1 unit sphere, we have $$||q(x)||_1 < |b_0| + \dots + \frac{1}{n+1}|b_n|$$ which implies that $$||q(x)||_{\infty} < (n+1)\left[ |b_0| + \dots + \frac{1}{n+1}|b_n|\right]$$ which means that it can not attain the maximum for the infinity norm, which means we suffer no loss of generality by ignoring them.
Thus, focusing on polynomials $1=\int_0^1 |p(x)| dx = |a_0| + \dots + \frac{1}{n+1}|a_n|$ will give us a strategy to actually attain the maximum for the infinity norm on the $L1$ unit sphere, because if any such polynomials exist, the upper bound assumes the form: $$||p||_{\infty} \le n+1$$
We get the most "bang for our buck" by trying to maximize the value associated with the $a_n$ term, since it gets reduced by a factor of $n+1$ when calculating the $L1$ norm, the largest reduction factor of any of the terms, while like the remaining terms it is left alone when calculating the $L\infty$ norm.
I.e. I claim that the monomial $(n+1)x^n$ assumes the maximum value of $n+1$ of $L\infty$ on the $L1$ unit sphere. This is clear because the $L\infty$ norm of $(n+1)x^n$ actually attains the upper bound given above for the $L\infty$ norm.
Therefore it follows that $C_N = n+1$. For the idea of the proof and how to generalize the computation, see here: https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf
EDIT: $|p(x)| = p(x)$ when $p(x) \ge 0$, $|p(x)|=-p(x)$ when $p(x)<0$, so |p(x)| is piecewise polynomial and the above argument suffers no loss of generality. Also note that the integral will be greatest (at least when we restrict the norm of the polynomials) when there are no changes of sign. So while this does suffer from some lack of details, it is still the right answer.
