It is well known that for a given polynomial $f \in \mathbb{Z}[x]$ the number of primes $p$ s.t. $f$ has a root modulo $p$ is infinite. In fact, one can even write down a formula for the density of such primes (e.g. D. Berend, Y. Bilu "Polynomials with roots modulo every integer"). 


Now suppose that $f = \sum\_{i=0}^d a\_ix^i$ is irreducible, the degree $d$ is fixed and all $a\_i \leq N^C$ for some constant $C$. Is it possible to prove the uniform bound (e.g. whith the constant depending only on $d$ and $C$ and $N$ large)
$$
\pi(N, f) = \Omega\_{d, C}(\frac{N}{\log N}),
$$
where $\pi(N, f)$ means the number of primes less than $N$ s.t. $f$ has a root modulo $p$? 


The aforementioned paper cites a general density formula due to Lagarias and Odlyzko which seem to provide such a bound assuming GRH. It's of course highly desirable to drop this assumption.



Uniform bound for the number primes $p$ s.t. a polynomial has a root modulo $p$
As Joel has pointed out this kind of question is delicate, and in fact even in the case $d=2$ no such result is known (for any fixed $C>0$). To see this, suppose that $p$ is a prime for which the quadratic character $\chi(n) = (\frac{n}{p})$ has a Siegel zero. That is there is a $\beta >1-\epsilon/\log p$ with $L(\beta,\chi)=0$. Of course we expect that there is no such character, but this is an important open problem. 


Now by the proof of the prime number theorem in arithmetic progressions (see Davenport's book on multiplicative number theory; this is called Page's theorem I think) we have 
$$
\psi(N,\chi) \sim - \frac{N^{\beta}}{\beta}, 
$$ 
where $\psi(N,\chi) =\sum\_{\ell \le N} \chi(\ell) \Lambda(\ell)$. Thus 
$$ 
\sum\_{\ell \le N, \chi(\ell)=1} \Lambda(\ell) \sim \frac 12 \Big( N- \frac{N^{\beta}}{\beta}\Big).
$$ 
If $N\le p^{1/C}$ then the RHS is $\ll \epsilon N$. 


To see the implication for your question, take $f(x)= x^2-p$ if $p\equiv 1\pmod 4$, and $f(x)=x^2+p$ if $p \equiv 3\pmod 4$. Then by quadratic reciprocity, if $f(x)\equiv 0 \pmod{\ell}$ has a solution, then $(\frac{\ell}{p})=1$. We have just shown that there are few primes $\ell$ below $N$ for which this happens. 


