Still, it seems worth recording what you know. I think at this point the question should be treated as "can we figure out anything?" – David E Speyer
Well, OK. Trying to figure out anything is exactly what we are now doing with Zachary Chase, and below is a small observation.
Let $\alpha>0$. Suppose that $n$ is even and we want to construct a polynomial $P_a(x)=\sum_{j=0}^n a_jx^j$ with $a_0=a_n=\alpha$, $a_j\in[0,1]$ for $1\le j\le n-1$ having roots $-1,-x,-x^2,\dots,-x^m$ where $x\in(0,1)$ is some number close to $1$.
It is possible if and only if the convex set $$ E=\{(a_0,a_n,P_a(x^k), 0\le k\le m):\\ a_0,a_n\in \mathbb R, a_j\in[0,1], 1\le j\le n-1\}\subset \mathbb R^{m+3} $$ contains the point $z=(\alpha,\alpha,0,\dots,0)$. So suppose that it does not. Then there is a non-trivial linear functional $\psi$ on $\mathbb R^{m+1}$ that is non-negativeon $E-z$, i.e., we can find some coefficients, not all $0$ such that $$ u(a_0-\alpha)+v(a_n-\alpha)+\sum_{k=0}^m w_kP_a(x_k)\ge 0 $$ for every admissible choice of the coefficient vector $a$.
Introduce the polynomial $Q(z)=\sum_{k=0}^m w_k z^k$. Then the condition above can be rewritten as $$ -u\alpha-v\alpha+(u+Q(1))a_0+(v+Q(x^n))a_n+\sum_{j=1}^{n-1}(-1)^jQ(x^j)a_j\ge 0\,. $$ Now, since $a_0$ and $a_n$ are free to run over the entire real line, we must have $u+Q(1)=v+Q(x^n)=0$. As to the rest of the expression, its minimum equals $$ U=\sum_{j=1}^{n-1} \min[(-1)^jQ(x^j),0]\,. $$ Thus we must have $$ \alpha(Q(1)+Q(x^n))+U\ge 0 $$ for some not identically $0$ polynomial $Q$ of degree at most $m$.
Now, when $x$ is really close to $1$, the points $x_j$, $j=1,\dots,n-1$ make an almost equispaced net on the interval $I=[x^n,1]$. Let $\mu=|Q(z)|=\max_I|Q|$. Then, by Markov's inequality, $|Q'|\le\frac 2{|I|}m^2\mu$ on $I$, so there is an interval of length $\frac{|I|}{4m^2}$ containing $z$ on which $Q$ preserves sign and is at least $\mu/2$ in absolute value. Since the powers $x_j$ are separated by about $|I|/n$, this interval contains about $cn/m^2$ powers $x^j$ with odd and even $j$. Choosing the parity appropriately, we get $U\le -c\frac{n}{m^2}\mu$.
On the other hand, $Q(1)+Q(x^n)\le 2\mu$. Hence we run into a contradiction when $2\alpha m^2\le cn$.
The conclusion: For every fixed $\alpha\ge 1$ and even $n$, there exists a polynomial $P_a(x)$ with $a_0=a_n=\alpha$, $a_j\in[0,1]$ ($j=1,\dots,n-1$) having $m$ distinct roots on $[-1,0)$, provided that $m^2\le c\alpha^{-1} n$.
It means that the non-negativity of the coefficients doesn't impose any substantial additional restrictions on the number of roots compared to the boundedness alone and the whole issue is the discretization from $[0,1]$ to $\{0,1\}$.
Edit I'll continue dumping obvious observations here. We'll now construct a polynomial with coefficients $0$ and $1$ having about $\frac{\log^2 n}{\log\log n}$ roots. While by itself this lower bound is rather pathetic, it still means that one should, probably, concentrate for a while on driving the lower bound up rather than the upper bound down.
The key observation is that $\sqrt[3]2<\frac 43$ by Bernoulli, so $$ -1+\sqrt[3]{\frac 12}+\frac 12-\left(\frac 12\right)^3-\left(\frac 12\right)^9 \\ >-1+\frac 34+\frac 12-\frac 18-\frac 18=0\,. $$ The immediate conclusion is that if $a=(-1,-1,1,1,-1,-1,1,1,\dots)$ and $p_j$ is an increasing sequence of positive integers such that $p_{j+1}\ge 3p_j$ for all $j$, then the polynomial $$ 1+\sum_{k=1}^u a_kx^{p_k} $$ has at least about $u/2$ sign changes on $(0,1)$. To turn it into $0,1$ polynomial with roots on $(-1,0)$, we need also to ensure that the parity of $p_j$ agrees with the sign of $a_j$, of course.
Now fix the (large) target degree $n$. Let $q,u,v$ be positive integers. Let $I_k=[3\cdot 6^{k-1} q, 6^k q]$ ($k=1,\dots,u$).
We will now choose $uv$ pairwise distinct integers $p_{ij}$ ($i=1,\dots,v, j=1,\dots,u$) so that for fixed $i$ one has $p_{ij}\in I_j$ for all $j$ and $p_{ij}$ has appropriate parity to be used in the above construction. We will also require that every positive integer $p$ can be written as the sum of at most $v$ integers $p_{ij}$ in at most one way. To ensure that such choice is possible by the mindless "just-take-what-is-still-available" algorithm, it is enough to require that $(uv)^{2v-1}<q$.
Now just put $P_i(x)=1+\sum_{j=1}^u x^{p_{ij}}$ and $P(x)=\prod_{i=1}^v P_i(x)$. Then $P$ is a $0,1$ polynomial of degree $\le 6^u qv$ and about $uv/2$ roots. To keep the degree under $n$ and to satisfy the previous condition, we choose $q\approx\sqrt n$, $u\approx c\log n$, $v\approx c\frac{\log n}{\log\log n}$ with sufficiently small $c>0$.
To be completely honest, I should also ensure that the roots of different $P_i$ are different too, but with that much freedom in choosing $p_{ij}$ that is rather trivial, so I'll leave it as an exercise to the interested readers.
Edit 2: Some more "Mathematische Banalen". This time we will show the existence of a polynomial with coefficients $0,\pm 1$ having $c\frac{\sqrt n}{\log n}$ zeroes on $[0,1]$. This still falls a bit short of $\sqrt n$, but is way better than $n^{1/4}$ David found in the literature for this case.
We start with a few observations about such polynomials. First, if $P$ is such a polynomial of degree $n$, then $$ \int_0^1 |P(x)|\,dx\ge n^{-C\sqrt n} $$ for large $n$ unless $P$ is identically $0$.
Indeed, if we use the same polynomial $Q$ and the differential operators $D_\lambda$, we will see that applying $D_{\lambda_s}$ with various $\lambda_s\in[0,n]$ to $P$ at most $C\sqrt n$ times, we'll get a polynomial $\widetilde P$ in which the first non-zero coefficient is $\pm 1$ and the sum of the absolute value of all other coefficients is below $1/2$. Then $\int_0^1|\widetilde P(x)|\,dx\ge\frac 12\int_0^1 x^n\,dx=\frac 1{2(n+1)}$.
However, due to the Markov's inequality $\|P'\|_\infty\le 2n^2\|P\|_\infty$, we have that the $L^1$ and the $L^\infty$ norms of $P$ are equivalent up to a factor $Cn^2$. Moreover, $|P|\ge \frac 12\|P\|_\infty$ on an interval of length $cn^{-2}$ around the point of the maximum of $P$. Thus, we have $\|P'\|_1\le Cn^4\|P\|_1$ and $\|P\|_\infty\le Cn^2 \int_{[0,1]\setminus E}|P|$ for every set $E$ of meaure $|E|\le cn^{-2}$.
The first conclusion shows that every application of $D_\lambda$ increases the $L^1$ norm of a polynomial of degree $n$ at most $Cn^4$ times, which immediately implies our first observation.
Now consider all polynomials $R$ of degree $n$ with coefficients $0,1$ and for each of them kompute their first $m+1$ moments $\int_0^1 R(x)x^k\,dx$ ($k=0,\dots,m$). Those are numbers in $[-n-1,n+1]$. Using the pigeonhole principle, as usual, and shamelessly exploiting the fact that the difference set is exactly what we need, we find a polynomial $P$ with coefficients $0,\pm 1$ that is not identically $0$ but satisfies $$ \left|\int_0^1 P(x)x^k\,dx\right|\le 2(n+1)2^{-\frac nm}\,. $$ for all $k=0,\dots,m$.
Now assume that $P$ has only $u<m$ distinct roots $r_j$ on $[0,1]$ at which a crossing occurs. Consider $q(x)=\prod_j(x-r_j)$. That is a polynomial of degree $u<m$ with the sum of absolute values of its coefficients at most $2^m$. Thus $$ \left|\int_0^1 P(x)q(x)\,dx\right|\le 2(n+1)2^{-\frac nm}2^m\,. $$
On the other hand, $Pq$ preserves sign and $|q|\ge (c'n^{-2})^m$ outside a set $E\subset \mathbb R$ of measure $|E|<cn^{-2}$ (Cartan's lemma). Hence the LHS equals $$ \int_0^1|P||q|\ge\int_{[0,1]\setminus E}|P||q|\ge (c'n^{-2})^m cn^{-2}\|P\|_\infty\ge (c'n^{-2})^m cn^{-2} n^{-C\sqrt n}\,, $$ so if $$ (c'n^{-2})^m cn^{-2} n^{-C\sqrt n}>2(n+1)2^{-\frac nm}2^m\,, $$ which happens for $m=c\frac{\sqrt n}{\log n}$, we get a contradiction.