The answer is **yes**, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on Wikipedia.

The details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T_n)$ with a general subset $\mathcal{T} \subseteq \mathcal{L}(X)$.

**Proof.** By Baire's Theorem we can find an integer $m$ such that the set
\begin{align*}
B := \{x \in A: \, \|Tx\| \le m \text{ for all } T \in \mathcal{T}\}
\end{align*}
has non-empty interior within $A$. Thus, we can find a point $x_0 \in B$ and a real number $\varepsilon \in (0,1]$ such that each $x \in A$ which has distance at most $\varepsilon$ to $x_0$ is contained in $B$.

Now, let $y \in A$. The vector $z := x_0 + \frac{\varepsilon}{2}(y-x_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x_0$ is at most $\varepsilon$ since both $y$ and $x_0$ have norm at most $1$. Thus, $\|Tz\| \le m$ for all $T \in \mathcal{T}$. Since
\begin{align*}
y = \frac{2}{\varepsilon}(z - x_0) + x_0,
\end{align*}
we conclude that
\begin{align*}
\|Ty\| \le \frac{4m}{\varepsilon} + m
\end{align*}
for all $T \in \mathcal{T}$. This bound does not depend on $y$.