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 $\{0,1\}$.