This is basically repeating what Joe Silverman wrote, in a concrete example. (I do so, as a misunderstanding between asker and answerers seems to persist.)
It might be difficult (or impossible) to get a lower bound similar to the upper bound. Let us fix the interval $[0,1]$.
For $x^2$ the derivative is $0$ in $0$, and this polynomial is thus not admissible by your question.
For $(x+0.001)^2$ the derivative is $0.002$ in $0$, so very small, and this polynomial is admissible by your question.
For $(x+1)^2$ the derivative is $2$ in $0$, so not too small.
So, one would need to distinguish 2 and 3. With this example one could still guess this might be due to the fact that in 2 the polynomial is small at 0. But replacing $(x+0.001)^2$ by 2'. $(x+0.001)^2 + 1$ the minimum is not too small anymore and just looking at rough parameters like maximum and minimum on the interval the polynomials 2' and 3 are not that different. While they are very different with respect to what you are looking for.
Thus, if a bound of the form you are looking for exists, it seems it definitely has to take into account other/finer quantities than maximum/minimum and degree. (I know you do not restrict exclusively to this situation, so this is not a full answer; but is meant to provide some 'bound' on what one can hope for.)