You can duplicate the usual proof of Hardy type inequalities to the discrete case.
Suppose $\{q_n\}$ is an eventually 0 sequence (you can weaken this to $\lim_{n\to \infty} n^{1/2} q_n = 0$). Then by telescoping you have (all sums are over $n\geq 0$)
$$ \sum (n+1) q_{n+1}^2 - n q_{n}^2 = 0 $$
Rewrite as
$$ \sum q_n^2 = - \sum (n+1) (q_{n+1} + q_n) (q_{n+1} - q_n) $$
Take absolute values and apply Cauchy-Schwarz on the RHS
$$ \sum q_n^2 \leq \left( \sum (n+1)^2 |q_{n+1} - q_n|^2 \right)^{1/2} \left( \sum (q_{n+1} + q_n)^2 \right)^{1/2} $$
the second factor can be bounded by $2 ( \sum q_n^2)^{1/2}$. Cancelling and you get
$$ \sum q_n^2 \leq C \sum (n+1)^2 |q_{n+1} - q_n|^2 $$
Scott Armstrong's comment is similar. As long as $\lim_{n\to\infty} q_n = 0$ you have
$$ q_n = \sum_{k \geq n} q_k - q_{k+1} $$
then
$$ \sup_{n\geq 0} |q_n| \leq \sum_{n\geq 0} |q_n - q_{n+1}| $$
If you want to look at "Sobolev type" inequalities: they are all essentially based on the fundamental theorem of (discrete) calculus applied in various ways.
Finally: note that you can also do a scaling argument.
Let $\lambda$ be a positive integer.
Let $q^{(\lambda)}_{n}$ be such that
$$ q^{(\lambda)}_m = q_n \text{ if } m \in [\lambda n, \lambda (n+1))$$
This scaling preserves the $ \sum |q_{n+1} - q_n|^2$ semi-norm, but has $ \sum |q^{(\lambda)}_n|^2 = \lambda \sum |q_n|^2$, which immediately falsifies your proposed Poincaré inequality.
You see that the inclusion of weights in Hardy avoids this difficulty.
Best Answer
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 $y$.