It is a well-known fact that roots of a unitary polynomial of fixed degree are continuous in the following sens :
Continuity of the roots
For all $n \geqslant 1$, $d : x,y \mapsto \min_{\sigma \in \frak{S}_n}\max_{1 \leqslant k \leqslant n} |x_k - y_{\sigma(k)}|$ is a semi-distance on $\mathbb{C}^n$ that becomes a distance when going to the quotient $\mathbb{C}^n/\frak{S}_n$ where $\frak{S}_n$ acts on $\mathbb{C}^n$ by permuting coordinates. Moreover, the topology induced by the distance is the quotient topology. Then the following function is continuous, $$ f : \left\{\begin{array}{rcl} \mathbb{C}^n & \rightarrow & \mathbb{C}^n/\frak{S}_n \\ (x_0,\ldots,x_{n - 1}) & \mapsto & \displaystyle \textrm{roots of } T^n + \sum_{k = 0}^{n - 1} x_kT^k \end{array}\right.. $$ To see this, consider for example for all $1 \leqslant k \leqslant n$, $S_{k,n}$ the symmetric polynomial of degree $k$ and $n$ variables $(X_0,\ldots,X_{n - 1})$ and set, $$ g : \left\{\begin{array}{rcl} \mathbb{C}^n & \rightarrow & \mathbb{C}^n \\ (x_0,\ldots,x_{n - 1}) & \mapsto & ((-1)^{n - k}S_{k,n}(x_0,\ldots,x_{n - 1}))_{0 \leqslant k \leqslant n - 1} \end{array}\right.. $$ Then, notice that $g \circ f$ is the projection $\mathbb{C}^n \rightarrow \mathbb{C}^n/\frak{S}_n$ by Newton formulas. We can verify that $g$ is continuous, proper and surjective and the projection is continuous by definition of the quotient topology, thus $f$ is continuous.
No uniform continuity of the roots
We can see however that $f$ is not uniformly continuous, in particular when considering polynomials with a discriminant close to $0$. Indeed, consider for all $a \geqslant 0$ large, $$ P = X^2 + 2aX + a^2, \quad Q = X^2 + (2a + a^{-1/2})X + a^2 + a^{-1/2}. $$ Then, the distances between the coefficients of $P$ and $Q$ have order $a^{-1/2}$ which converges toward $0$ when $a \rightarrow +\infty$ but their roots are getting farther and farther. Indeed, the roots of $P$ are $(-a,-a)$ and the ones of $Q$ are, $$ \frac{1}{2}(-2a - a^{-1/2} \pm \sqrt{(2a + a^{-1/2})^2 - 4(a^2 + a^{-1/2})}) = -a \pm \frac{1}{2}\sqrt{4a^{1/2} + \mathrm{O}(1)} + \mathrm{O}(1) = -a \pm a^{1/4} + \mathrm{O}(1). $$ Therefore, the distance bewteen the roots of $P$ and the roots of $Q$ grows toward infinity with order $a^{1/4}$, proving $f$ is not uniformly continuous (at least when $n = 2$, the same proof holds otherwise).
Question
My question is : what happens when only the constant term varies ? In other words, it is true that for all $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x = (x_0,x_1,\ldots,x_{n - 1})$ and $y = (y_0,x_1,\ldots,x_{n - 1})$ with $|x_0 - y_0| \leqslant \delta$, then $d(f(x),f(y)) \leqslant \varepsilon$ ?
Positive answer when $n = 2$
This is, here, true when $n = 2$. Indeed, consider some $\delta > 0$ and $(x_0,y_0,x_1)$ whith $|x_0 - y_0| \leqslant \delta$. Then, the roots of $P = X^2 + x_1X + x_0$ and $Q = X^2 + x_1X + y_0$ are respectively $\frac{1}{2}\left(-x_1 \pm \sqrt{x_1^2 - 4x_0}\right)$ and $\frac{1}{2}\left(-x_1 \pm \sqrt{x_1^2 - 4y_0}\right)$. Therefore, the distance between $f(x_0,x_1)$ and $f(y_0,y_1)$ is $D = \left|\sqrt{x_1^2 - 4x_0} \pm \sqrt{x_1^2 - 4y_0}\right|$ for the best choice of complex square roots.
First case : $|x_1^2 - 4x_0| \leqslant 6\delta$. Then, $|x_1^2 - 4y_0| \leqslant |x_1^2 - 4x_0| + |4x_0 - 4y_0| \leqslant 10\delta$ so $$ D \leqslant \left|\sqrt{x_1^2 - 4x_0} \pm \sqrt{x_1^2 - 4y_0}\right| \leqslant \left|\sqrt{x_1^2 - 4x_0}\right| + \left|\sqrt{x_1^2 - 4y_0}\right| \leqslant (\sqrt{6} + \sqrt{10})\sqrt{\delta}. $$
Second case : $x_1^2 - 4x_0 > 6\delta$. Then $|x_1^2 - 4y_0 - x_1^2 + 4x_0| = |4y_0 - 4x_0| \leqslant 4\delta$. Let $U$ be the disk of center $x_1^2 - 4x_0$ and radius $5\delta$. We have $x_1^2 - 4x_0 \in U$ and $x_1^2 - 4y_0 \in U$, and $U$ is a simply connected open subset of $\mathbb{C}^*$, meaning we can define a holomorphic square root $\sqrt{\cdot}$ on $U$. Moreover, any element of $U$ has a module $\geqslant \delta$.
Using this square root and Taylor inequality at order $1$, $$ D \leqslant \left|\sqrt{x_1^2 - 4x_0} - \sqrt{x_1^2 - 4y_0}\right| \leqslant \sup_U \left|\frac{1}{2\sqrt{\cdot}}\right|\left|x_1^2 - 4x_0 - x_1^2 + 4y_0\right| \leqslant \frac{1}{2\sqrt{\delta}}4\delta = 2\sqrt{\delta}. $$ In both case, we have $D \leqslant (\sqrt{6} + \sqrt{10})\sqrt{\delta}$. I believe in degree $n$, we always have $D = \mathrm{O}(\delta^{1/n})$ where the $\mathrm{O}$ is uniform, but I haven't been able to prove it. The proof for $n = 2$ suggests that we have to distinguish the cases were two roots are close to each other (small discriminant), and the other cases.
I found an answer so I put it here, the answer is yes !
The idea of the proof is that you can write any unitary polynomial $P$ as $\prod_{k = 1}^n (x - r_k)$ thus when $|P(x)| \leqslant 1$, there exists at least one $k$ such that $|x - r_k| \leqslant 1$. It means that the reciprocal image $A$ of the closed unit disk by $P$ is included in a union of $n$ unit disks. In particular, each connected component of $A$ has a diameter less than or equal to $2n$.
By continuity of the roots of a polynomial, then number of roots of $P$ in each connected component of $A$ equals the number of roots of $P - 1$ in this same connected component (all roots are always counted with multiplicities). It implies that the distance between the roots of $P$ and the roots of $P - 1$ is less than or equal to $2m$.
Now, let $x = (x_0,x_1,\ldots,x_{n - 1})$ and $y = (y_0,x_1,\ldots,x_{n - 1})$ such that $0 < |x_0 - y_0| \leqslant \delta$ for some $\delta > 0$, let $\omega$ be a $n$-th root of $x_0 - y_0$ and set $P(X) = \frac{1}{x_0 - y_0}\left(\omega^nX^n + \sum_{k = 0}^{n - 1} x_k\omega^kX^k\right)$, which is unitary. We can use the previous fact on $P$ and notice that the roots of $P$ are the roots of $X^n + \sum_{k = 0}^{n - 1} x_kX^k$ divided by $\omega$ and the roots of $P - 1$ are the roots of $X^n + \sum_{k = 1}^{n - 1} x_kX^k + y_0$ divided by $\omega$. Therefore, $d(f(x),f(y)) = |\omega| \cdot \textrm{distance bewteen the roots of $P$ and the roots of $P - 1$} \leqslant 2n\delta^{1/n}$. It proves the wanted result.