To add to T. Amdeberhan's answer, the inequality $$|c_{n}|\leq1/n$$ is actually sharp since $nz+z^{n}$ is univalent in the open unit disk. Indeed, for $z\neq u$, $$nz+z^{n}=nu+u^{n}\quad\Longleftrightarrow\quad -n=z^{n-1}+z^{n-2}u+\cdots+u^{n-1},$$ and the sum on the right-hand side has modulus less than $n$ for $z$ and $u$ in the open unit disk.

Moreover, concerning the other coefficients, the following result is proved in

Suffridge, T. J. On univalent polynomials. J. London Math. Soc. 44, 1969, 496-504.

Let $$C_{k,j}=\frac{n-k+1}{n}\frac{\sin jk\pi/(n+1)}{\sin j\pi/(n+1)},\qquad k=2,\ldots,n,\quad j=1,2.$$ Then, for univalent polynomials with real coefficients and such that $c_{n}=1/n$, the coefficients $c_{k}$ satisfy the sharp inequalities $$|c_{k}|\leq |C_{k,1}|, $$ and for univalent polynomials with real coefficients and such that $c_{n}=-1/n$, $$ |c_{k}|\leq\begin{cases} |C_{k,1}|\quad(n\text{ even})\\ |C_{k,2}|\quad(n\text{ odd}). \end{cases} $$

Concerning the other coefficients $c_k$, Branan [1] showed that $(k+1)c_{k+1}=(n-k)\bar c_{n-k}$ is a necessary condition for univalence in $|z|<1$ if $c_1=1$ and $c_n=1/n$.

For univalent polynomials with real coefficients and such that $c_{n}=1/n$, Suffridge [2] has proved that the coefficients $c_{k}$ satisfy the inequalities $$|c_{k}|\leq |C_{k,1}|, $$ and for univalent polynomials with real coefficients and such that $c_{n}=-1/n$, $$ |c_{k}|\leq\begin{cases} |C_{k,1}|\quad(n\text{ even})\\ |C_{k,2}|\quad(n\text{ odd}), \end{cases} $$ where $$C_{k,j}=\frac{n-k+1}{n}\frac{\sin jk\pi/(n+1)}{\sin j\pi/(n+1)},\qquad k=2,\ldots,n,\quad j=1,\ldots,n.$$ For each $n=1,2,\ldots$, and $1\leq j\leq n$, the polynomials $$P(z)=\sum_{k=1}^n C_{k,j}z^k$$ are univalent in $|z|<1$, from which follows that the above inequalities are sharp.

[1] D. A. Brannan, Coefficient regions for univalent polynomials of small degree, Mathematika, 14 (1967), 165-169.

[2] T. J. Suffridge, On univalent polynomials. J. London Math. Soc. 44 (1969), 496-504.