Let $f \in \mathcal{H}(\mathbb{C} \setminus \{0\})$ be and assume that
$$ |f(z)| \leq |\log|z|| + 1, \quad z \in \mathbb{C} \setminus \{0\} $$
I have to prove that $f$ is constant. My attempt is next proof. Consider $g(z) = zf(z)$. By hypothesis $$ |z||f(z)| \leq |z||\log|z|| + |z|, $$ so $\lim_{z \to 0} g(z) = 0$ and $g$ is a entire function, by Riemann's theorem. By Cauchy integral formula $$ |g^{(n)}(0)|\leq \dfrac{M}{r^{n-1}}(r\log(r) + r). $$
Is easy to see that for all $n > 2$, $g^{(n)}(0) = 0$ because the right hand part has limit zero when $r \to \infty$. For $n = 1$, we obtain the same result taking limit $r \to 0$. So Taylor expansion $(g(0) = 0$) $$ g(z) = az^2, \quad a \in \mathbb{C}. $$ For all $z \neq 0$, $f(z) = az$. By hypothesis, if $z \neq 0$ $$ |a| \leq \dfrac{|\log|z|| + 1}{|z|}. $$ Taking $|z| \to \infty$ we obtain $|a| = 0$, so $f \equiv 0$.
Is correct that argument? It seems a little strange because the exercise says that $f$ is constant, not necessarily zero in every point.
There is an error in your application of the Cauchy integral formula.
The correct estimate is $$ |g^{(n)}(0)| \le \frac{n!}{r^n} \max_{|z|=r} |g(z)| \le n! \frac{|\log(r)| + 1}{r^{n-1}} $$ which implies $g^{(n)}(0) = 0$ for $n \ge 2$, so that $g$ is linear, and consequently, $f$ is constant.
Finally, setting $z=1$ shows that $f(z) \equiv a$ with $|a| \le 1$.