UPDATE I was right the first time after all.
I claim that the degree $n$ monic polynomial with minimal $L_1$ norm is $\prod_{k=1}^n (x-\cos^2(\pi k/(2n+2)))$.
By standard compactness arguments, there is such a minimal polynomial, although it is not clear whether or not it is unique. Let $f$ be a polynomial which achieves the minimum.
Lemma: The polynomial $f$ has $n$ distinct roots in the interval $(0,1)$.
Proof: Suppose to the contrary that $f$ had fewer roots. Let $\rho_1$, $\rho_2$, ..., $\rho_m$ be those roots of $f$ in $(0,1)$ that have odd multiplicity. Then $f$ has the same, or opposite, sign as $g(x) := \prod(x-\rho_i)$ everywhere on $[0,1]$. For $\epsilon$ sufficiently small, and the right choice of sign, $| f \pm \epsilon g(x) |_1 < |f|_1$. $\square$
Let the roots of $f$ be $0 < \rho_1 < \rho_2 < \cdots < \rho_n <1$, and formally define $\rho_0=0$ and $\rho_{n+1}=1$ for convenience.
As we learn in any calculus class, in order to minimize a function, we shold figure out how it changes when the input is perturbed. Let $g$ be any polynomial of degree $. For small $\epsilon$, the implicit function theorem tells us that $f + \epsilon g$ has $n$ roots in $(0,1)$, given by smooth functions $\rho_i(\epsilon)$. (We are using the lemma to know that $f'(\rho_k) \neq 0$, so that we may apply the implicit function theorem.) We set $\rho_0(\epsilon)=0$ and $\rho_{n+1}(\epsilon) = 1$.
Then $|f+\epsilon g|_1 = \sum_{k=0}^n (-1)^{n-k} \int_{\rho_k(\epsilon)}^{\rho_{k+1}(\epsilon)} f+\epsilon g$ $ = \sum_{k=0}^n (-1)^{n-k} \left( \int_{\rho_k}^{\rho_{k+1}} f+\epsilon g + \int_{\rho_k}^{\rho_k(\epsilon)} f+\epsilon g - \int_{\rho_{k+1}}^{\rho_{k+1}(\epsilon)} f+\epsilon g \right)$ $=|f|_1 + \epsilon \sum_{k=0}^n (-1)^{n-k} \int_{\rho_k}^{\rho_{k+1}} g + \mbox{error}$ where the error comes from the latter two integrals.
Let's look at these integrals. Since $\rho_k$ is a smooth function, we have $\rho_k(\epsilon)= \rho_k + O(\epsilon)$, so the width of the interval is $O(\epsilon)$. Also, $f(\rho_k) =0$ and $f$ is smooth so $f(x) = O(\epsilon)$ for $x \in (\rho_k, \rho_k+O(\epsilon))$. Finally, $g$ is bounded, so $g=O(1)$. So our error terms are all of the form $\int_{\rho_k}^{\rho_k + O(\epsilon)} \left( O(\epsilon) + \epsilon O(1) \right) = O(\epsilon^2).$ So $|f+\epsilon g|_1 = |f|_1 + \epsilon \sum_{k=0}^n (-1)^{n-k} \int_{\rho_k}^{\rho_{k+1}} g + O(\epsilon^2).$
We see that the directional derivative of $| \ |_1$ in direction $g$ exists. So, if $f$ is to be a minimum of $| \ |_1$, then we must have
Criterion: For every polynomial $g$ of degree $< n$, we have $\sum_{k=0}^n (-1)^{n-k} \int_{\rho_k}^{\rho_{k+1}} g=0$
Uniqueness of solution: I will first show that there is a unique $n$-tuple $(\rho_1, \ldots, \rho_n) \in [0,1]^n$ obeying the criterion. Observe that the criterion is linear in $g$, so it is enough to check the criterion for the monomials of degree $. To keep notation simple, I'll restrict myself to the case $n=2m$. It is convenient to rename $(\rho_1, \ldots, \rho_{2m})$ as $(\alpha_1, \beta_1, \alpha_2, \beta_2, \ldots, \beta_m)$.
Applying the Criterion with $g=x^{k-1}$ we see that, for $1 \leq k \leq n$, we have $\frac{1}{k} \left( \alpha_1^k - (\beta_1^k-\alpha_1^k) + (\alpha_2^k-\beta_1^k) - \cdots +(1-\beta_m^k) \right)=0.$ Equivalently $2 \sum \frac{\alpha_i^k}{k} + \frac{1}{k} = 2 \sum \frac{\beta_i^k}{k} \ \mathrm{for} \ 1 \leq k \leq n.$
Now comes the part I thought was clever. Organize the above $n$ equations into a generating function: $\sum \frac{\alpha_i^k x^k}{k} + \frac{1}{2} \sum \frac{x^k}{k} = \sum \frac{\beta_i^k x^k}{k} + O(x^{n+1}).$ This gives $\sum \log \frac{1}{1-\alpha_i x} + \frac{1}{2} \sum \log \frac{1}{1-x} = \sum \log \frac{1}{1-\beta_i x} + O(x^{n+1})$ or $\prod (1-\alpha_i x) \sqrt{1-x} = \prod (1-\beta_i x) + O(x^{n+1}).$ Set $\prod(1-\alpha_i x) = 1+a_1 x + a_2 x^2 + \cdots + a_m x^m$ and $\prod(1-\beta_i x) = 1+b_1 x + b_2 x^2 + \cdots + b_m x^m$. So $\left( 1+a_1 x + a_2 x^2 + \cdots + a_m x^m \right) \sqrt{1-x} = \left( 1+b_1 x + b_2 x^2 + \cdots + b_m x^m \right) + O(x^{n+1}).$
Writing our this power series an equating coefficients of $x^k$ for $k\leq n$, we get linear equations for the $a_i$ and $b_j$. In particular, the set of solutions is a linear space and it shouldn't be too hard (I didn't actually check this) to see that it is a zero dimensional space. So, subject to the linear algebra problem, the solution is unique.
The key practical point is that computers solve linear equations very quickly. So I had Mathematica go and solve these equations for me for several values of $n$. For $n=10$, I got $a(x) = \frac{1}{1024} \left( 1024 - 2816 x + 2816 x^2 - 1232 x^3 + 220 x^4 - 11 x^5 \right)$ $b(x) = \frac{1}{1024} \left( 1024 - 2304 x + 1792 x^2 - 560 x^3 + 60 x^4 - x^5 \right).$ So I typed $1024, -2816, 2816, -1232, 220, -11$ and $1024, -2304, 1792, -560, 60, -1$ into Sloane's encyclopedia and immediately learned that these were coefficients of Chebyshev polynomials of the first and second kind. It took a little while to get all the indexing issues right, but I eventually realized that this meant my solution was $f(x) = \prod_{k=1}^n (x-\cos^2(\pi k/(2n+2))).$
Once we know that the solution is unique, it is enough to check that $f$ obeys the Criterion. Let $h$ be the function $[0,1] \to \{ -1 , 1 \}$ which switches sign at the numbers $\cos^2(\pi k/(2n+2))$. We want to show:
Claim For any polynomial $g$ of degree $, we have $\int_0^1 h(x) g(x) dx=0.$
The form of $h$ suggests putting $x=\cos^2 t$, so we want $\int_0^{\pi/2} h(\cos^2 t) g(\cos^2 t) (2 \sin t \cos t) dt=0.$ Now, $h(\cos^2 t)$ switches signs at $k \pi/(2n+2)$. Let $\sigma$ be the square wave function: $\sigma(u)$ is $\pm 1$ and switches signs at integer multiples of $\pi$. So we want $\int_{0}^{\pi/2} \sigma( (2n+2) t) g((\cos (2t)+1)/2) \sin(2t) dt=0.$ Setting $u=2 t$, and noticing that the integrand is an even function, we get $\int_{-\pi}^{\pi} \sigma( (n+1) u) g((\cos u+1)/2) \sin u \ du=0.$
Also, if $g$ is an polynomial of degree $, then the polynomial $\hat{g}(w) := g((u+1)/2)$ is also has degree $. So our goal is to show:
For any polynomial $\hat{g}$ of degree $, we have $\int_{-\pi}^{\pi} \sigma((n+1) u) \hat{g}(\cos u) \sin u \ du=0$.
It is enough to check this for a basis of polynomials $\hat{g}_0$, ..., $\hat{g}_{n-1}$; we use the Chebyshev polynomials, so that $\hat{g}_m(\cos u) = \cos (mu)$. So our goal can be rewritten as showing that $\int_{- \pi}^{\pi} \sigma((n+1) u) \cos (mu) \sin u \ du=0 \ \mathrm{for} \ 0 \leq m < n$ or that $\frac{1}{2} \int_{- \pi}^{\pi} \sigma((n+1) u) \left( \sin ((m+1) u) - \sin((m-1) u) \right) du \ \mathrm{for} \ 0 \leq m < n.$
But $\sigma(v)$ has a well known Fourier series $\sin(v) + \frac{1}{3} \sin 3 v + \frac{1}{5} \sin (5v) + \cdots$, so $\sigma((n+1)u)$ has a Fourier series starting $\sin((n+1)u) + \cdots$. We see that $\sigma((n+1) u)$ is orthogonal to the functions $\sin (ru)$ for $r \leq n$, which immediately establishes the required claim. QED