*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][1], $|Q'|\le\frac 2{|I|}m^2\mu$ 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.




  [1]: https://en.wikipedia.org/wiki/Markov_brothers%27_inequality