In this case, we can give an easy estimate on the tail probability of $T_1$. Notice that
$$ \{T_1 = 2n+1\} = \{S_1 \leq 0, \cdots, S_{2n-1} \leq 0, S_{2n} = 0, S_{2n+1} = 1\}. $$
Using one of the equivalent characterization of Catalan number, we can explicitly compute the probability of this event as
$$ \Bbb{P}(T_1 = 2n+1) = \frac{1}{2^{2n+1}}C_n = \frac{1}{2^{2n+1}(n+1)} \binom{2n}{n}. $$
From this, we explicitly compute the probability generating function of $T_1$ by
$$ |z| < 1 \quad \Rightarrow \quad \mathbb{E}[z^{T_1}] = \sum_{n=0}^{\infty} z^n \mathbb{P}(T_1 = n) = \sum_{n=0}^{\infty} \frac{C_n}{2^{2n+1}} z^{2n+1} = \frac{1-\sqrt{1-z^2}}{z}. $$
Letting $z \to 1^-$ shows that, by the monotone convergence theorem,
$$ \mathbb{P}(T_1 < \infty) = \lim_{z \to 1^-} \mathbb{E}[z^{T_1}] = \lim_{z \to 1^-} \frac{1-\sqrt{1-z^2}}{z} = 1. $$
Therefore $\Bbb{P}(T_1 = \infty) = 0$.
Addendum. Using this, we can also show that
$$ \mathbb{E}[T_1 z^{T_1}] = \frac{1-\sqrt{1-z^2}}{z\sqrt{1-z^2}}, $$
and so, $T_1$ infinite expectation $\Bbb{E}[T_1] = \infty$.
Here's an easy way to do it:
Let $E X = -\gamma < 0$; I'm gonna assume that $|X| \leq 1$, but this works even without this assumption. By Azuma's inequality, we have $$P(S_n \geq -\gamma \cdot n/2) \leq \exp\left( -\frac{\gamma^2 n}{8}\right)\,.$$
Thus there is some $M$ so that $$\sum_{n \geq M} P(S_n \geq -\gamma \cdot n/2) < 1\,.$$
Let $A_M$ denote the event that $ \{S_n \leq - \gamma \cdot n/2 \text{ for all }n \geq M\}$. Then the above shows that $P(A_M) > 0$. But on the event $A_M$, the expectation you are interested in is bounded below.
Best Answer
Let $A_n$ be the event "$0 \leq S_n \leq x$ and $\min_{0 \leq k \leq n} S_k \geq 0$" appearing in the question. Then the sum in question can be rewritten as $$ \lim_{t \to \infty} E[\#(n \leq t : A_n)]. $$
Choose $\epsilon > 0$ such that $P(X \leq -\epsilon) = \delta > 0$. Then if it ever happens that $S_n \leq x$, there is a positive probability that the random walk will dip below 0 in bounded time. Namely, setting $c = \lfloor x/\epsilon \rfloor + 1$, we have $$ P(S_{n+c} < 0) \geq P(X_{n+1}, \dots, X_{n+c} < -\epsilon) = \delta^c > 0. $$ It follows that conditional on $A_n$, there is a probability $\geq \delta^c$ that $A_m$ will never occur for $m \geq n+c$.
Now fix some $N \in \mathbb N$ (to avoid worrying about convergence issues), and let $k \in \mathbb N$ vary. We claim that $$ P(\#(n \leq N : A_n) \geq k \cdot c) \leq (1 - \delta^c)^k. $$ Proof: Induct on $k$, with the base case $k = 0$ being trivial. If $A_n$ has already occurred $k \cdot c$ times, the probability that it will occur at least $c$ more times is at most $1 - \delta^c$.
Since $(1 - \delta^c)^k$ decreases exponentially, we get a constant upper bound on $E[\#(n \leq N : A_n)]$. This doesn't depend on $N$, so it gives an upper bound on $\lim_{t \to \infty} E[\#(n \leq t : A_n)]$ as well.