I'm not sure I understand your question. $M(f)\ge1$ for all $f$, and if you take $f(x)=x^d+x^{d-1}+\cdots+1$, them $M(f)=1$, since $f$ is a product of cyclotomic polynomials. So you can't improve on the lower bound of $1$. If you want to exclude polynomials containing cyclotomic factors, then it's not clear that assuming the largest coefficient has absolute value $1$ will help you. For polynomials with integer coefficients, there is a bound of the form $M(f) \ge C (\log d)/(\log\log d)^{-3}$ due to Dobrowolski, but you probably already know that. OTOH, if you're really going to allow arbitrary complex coefficients, then I'm surprised you can get a bound even for non-cyclotomics.
ADDENDUM for non-monic polynomials
Let $|f|$ denote the magnitude of the largest coefficient of $f\in\mathbb C[x]$. There's a non-trivial upper bound for the Mahler measure
$$ M(f) \le |f|\cdot (\deg f + 1)^{1/2}, $$
see for example Fundamentals of Diophantine Geometry, Serge Lang, Springer, Chapter 3, Theorem 2.8. (I realize you're asking for a lower bound.) The lower bound provided by the same theorem decays exponentially,
$$ M(f) \ge 2^{-\deg f}|f|. $$
It's not clear you can do too much better than that. For example,
$$f(x) = \frac{(x+1)^{2d}}{\binom{2d}{d}}$$ has largest coefficient $1$ and,
since $M((x+1)^d)=M(x+1)^d=1$, it has Mahler measure
$$ M(f) = \frac{1}{\binom{2d}{d}} \asymp \frac{\sqrt{d}}{4^d} \asymp \frac{\sqrt{\deg f}}{2^{\deg f}}.$$
In any case, you can't hope for better than an exponential lower bound.
