For $t$ fixed, the count is proportional to $\lambda^n$, where $\lambda = 2 \cos \frac\pi{2t+2}$ is the principal eigenvalue of the adjacency matrix of the path with $2t+1$ vertices. The all-positive (Perron-Frobenius) eigenvector corresponding to $\lambda$ is

$$\bigg(\sin \frac{\pi}{2t+2}, \sin \frac{2\pi}{2t+2},\sin \frac{2\pi}{2t+2},\dots,sin \frac{(2t+1)\pi}{2t+2}\bigg).$$

Since $-\lambda$ is also an eigenvalue, the stable behavior of the distribution of endpoints of paths which stay in $[-t,t]$ is an oscillation between the odd entries

$$\bigg(\sin \frac{\pi}{2t+2}, 0,\sin \frac{3\pi}{2t+2},0,\dots,\sin \frac{(2t-1)\pi}{2t+2},0,\sin \frac{(2t+1)\pi}{2t+2}\bigg).$$
and even entries
$$\bigg(0,\sin \frac{2\pi}{2t+2}, 0,\sin \frac{4\pi}{2t+2},0,\cdots ,0,\sin \frac{2t\pi}{2t+2},0\bigg).$$

The exact count of paths staying in $[-t,t]$ is a sum of signed binomial coefficients.

The number of paths from $0$ to $i$ is 0 if $n \not \equiv i ~\mod 2$, and $n \choose (n\pm i)/2$ when $n \equiv i ~\mod 2$.

The number of paths which never leave $[-t,t]$ from $0$ to $i \in [-t,t]$ with $n \equiv i ~\mod 2$ is

$$ \sum_{j\in \mathbb Z} (-1)^j {n\choose (n +i)/2 + j(t+1)}$$

by the reflection principle applied to the group of isometries of $\mathbb R$ generated by reflecting about $t+1$ and $-t-1$.

If you sum over all $i \in [-t,t]$, then when $n$ is even, you get a signed sum of binomial coefficients with $t+1$ positive signs in a row alternating with $t+1$ negative signs in a row. If $n$ is odd, then you get $t$ positive signs in a row, skip a term (give it a coefficient of $0$ instead of $\pm 1$), then $t$ negative signs in a row, skip a term, etc.

For example, for $n=100, t=2,$ the number of paths is

$$ ... +{100\choose 43} + {100\choose 44} + {100 \choose 45} - {100 \choose 46} - {100 \choose 47} - {100\choose 48} + {100\choose 49} + {100 \choose 50} + {100\choose 51} - ...$$

For $n=101, t=2,$ the number of paths is

$$ ... +{101\choose 44} + {101\choose 45} - {101\choose 47} - {101 \choose 48} + {101\choose 50} + {101\choose 51} - {101\choose 53} - {101\choose 54} + ...$$

These can be summed using the techniques in the answers to the Binomial distribution parity question.

A lot more can be said when $t$ varies, but the answers are more complicated. For $t$ slowly increasing, as $c\sqrt[3]n$, there is enough time for the distribution to stabilize (for each parity) at a given value of $t$, since the ratio between the magnitudes of the largest two eigenvalues and the magnitudes of the next two is about $1+c/t^2$, and the principal eigenvectors have a small $L^1$ distance for adjacent values of $t$. You should pick up a constant factor for each transition. In other words, the number of paths when you spend at least $n_t \gt c t^2$ steps at a given $t$ should be

$$C \prod_{t \le t_{max}} (2 \cos \frac{\pi}{2t+2})^{n_t}$$

where $C$ is between some functions $f_{lower}(t_{max}) \lt C \lt f_{upper}(t_{max})$ which does not depend on the values of $n_t$. I don't think the $n_t \gt c t^2$ condition is sharp for this behavior. Something like $n_t \gt c t^2/\log t$ should work, too. The geometry of the eigenvectors for adjacent values of $t$ lets you estimate $f_{lower}$ and $f_{upper}$.

For $t$ more rapidly increasing, different behaviors occur. By the law of the iterated logarithm, if $t$ increases as $t(n) = \sqrt {(2-\epsilon) n \log\log n},$ random paths will almost surely violate the constraint. I think there are precise versions of the law of the iterated logarithm which may tell you when a positive proportion of random paths do not violate the constraint. I would guess that if $t(n) = \sqrt{(2+\epsilon) n \log\log n}$ then a positive percentage of random paths won't violate the constraint.