There is a simple recursion formula to calculate $f(n,t)$:
$$
f(n,t) =
\begin{cases}
0 & \text{if $t<0$} \\[1ex]
1 & \text{if $t=0$} \\[1ex]
\frac{1}{n} & \text{if $t=1$} \\[1ex]
\frac{n+1}{n}f(n,t-1)-\frac{1}{n}f(n,t-1-n)) & \text{if $t>1$}
\end{cases}
$$

Proof: Obvious for the cases up to $t=1$.
For $t>1$ consider that $f(n,t)$ is the sum of the probabilities of
all paths to $t$ (a path to $t$ being a sequence of integers that sum up to $t$).

All paths to $t$ can be constructed from a path to $t-1$ by exactly one of the following

(1) prepending a $1$ (all paths to $t-1$ can be used)

(2) increasing the first number (only paths not starting with $n$ can be used)

The probability of a path of type (1) is $\frac{1}{n}$ of the probability of the original path, so the probabilities of the paths of type (1) sum up
to $\frac{1}{n}f(n,t-1)$.

The probability of a path of type (2) equals the probability of the original path. So the probability of paths of type (2) equals $f(n,t-1)$ minus the sum of the probabilities of paths to $t-1$ that start with $n$.
The latter sum equals $\frac{1}{n}f(n,t-(n+1))$ (first roll $n$ and then take a path to $t-1-n$) $\square$

It is interesting to look at the first 30 values of $f(6,t)$. First we see a steep increase due to the increasing possibilities. Then at $t=7$ a sudden drop, because we lose the possibility to roll it in one turn. Then it evens out very quickly. We see a small bump at about $t=13$ as an echo to the first drop.

An alternative recursion formula is

$$
f(n,t) =
\begin{cases}
0 & \text{if $t<0$} \\[1ex]
1 & \text{if $t=0$} \\[1ex]
\frac{1}{n}\sum_{i=1}^n{f(n,t-i)} & \text{if $t>0$}
\end{cases}
$$

So starting from $t=1$ each value is the average of the preceeding $n$ values. This can be shown in a similar manner, considering each possibility for the first roll separately.

From this, it is easy to see that the values increase from $t=1$ to $t=n$ and after that never reach the maximum value at $t=n$ again.

Regarding the moving window of the average, it can be seen that the range of fluctuations between the values is dampened by at least a factor $n$ after each $n$ values. Thus the sequence converges. The probability to hit some high $t$ must be close to the average density of the paths, which is reciprocal to the average dice value. Thus the limit of $f(n,t)$ must be $2/(n+1)$