This would indeed be related to the coupon collector’s problem if you were interested in the actual average, but finding the “unweighted” average you want is a simpler combinatorial problem.
First, note that the transformation from $\vec T$ to $\vec\Delta$ is linear and bijective. Thus, distinct values of $\vec T$ correspond to distinct values of $\vec\Delta$, and the transformation commutes with the average; so we can average $\vec T$ and only then transform to $\vec\Delta$.
Clearly $t_1$ is always $1$. We have $2\le t_2\le h+1$, since the second distinct letter must appear at the latest after all $h$ copies of the first letter. Likewise, $t_2+1\le t_3\le2h+1$, since the third distinct letter must appear at the latest after all $2h$ copies of the first two distinct letters. Thus, the number of distinct values of $\vec T$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}1 =\frac12 h(3h-1)\;,
$$
the sum of $t_2$ over all distinct values of $\vec T$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}t_2 =\frac23h(h^2+3h-1)\;,
$$
and the sum of $t_3$ over all distinct values of $\vec T$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}t_3 =\frac16h(11h^2+12h-5)\;.
$$
So asymptotically for $h\to\infty$, we have $\vec{\overline T} \sim\left(1,\frac{\frac23h^3}{\frac32h^2},\frac{\frac{11}6h^3}{\frac32h^2}\right)=\left(1,\frac49h,\frac{11}9h\right)$, and thus
$\vec{\overline\Delta}\sim(1,\frac49h,\frac79h)$, in agreement with your results.
We can do the same thing for $n=4$; it's just slightly more tedious. As before, we have $t_3+1\le t_4\le3h+1$, so the number of distinct values of $\vec T$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}\sum_{t_4=t_3+1}^{3h+1}1 =\frac13h(8h^2-6h+1)\;,
$$
the sum of $t_2$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}\sum_{t_4=t_3+1}^{3h+1}t_2 =\frac1{24}h(27h^3+74h^2-63h+10)\;,
$$
the sum of $t_3$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}\sum_{t_4=t_3+1}^{3h+1}t_3 =\frac1{12}h(35h^3+30h^2-35h+6)\;,
$$
and the sum of $t_4$ is
$$
\sum_{t_2=2}^{h+1}\sum_{t_3=t_2+1}^{2h+1}\sum_{t_4=t_3+1}^{3h+1}t_4 =\frac1{24}h(131h^3+22h^2-71h+14)\;.
$$
Thus, asymptotically for $h\to\infty$ we have $\vec{\overline T}\sim\left(1,\frac{\frac{27}{24}h^4}{\frac83h^3},\frac{\frac{35}{12}h^4}{\frac83h^3},\frac{\frac{131}{24}h^4}{\frac83h^3}\right)=\left(1,\frac{27}{64}h,\frac{35}{32}h,\frac{131}{64}h\right)$, corresponding to $\vec{\overline\Delta}\sim\left(1,\frac{27}{64}h,\frac{43}{64}h,\frac{61}{64}h\right)$, also in agreement with your results.
Best Answer
I do not think there is a nice formula, but there is a generating function solution.
Let $E_n(x)=\sum_{j=0}^n\frac{x^j}{j!}$ be the partial exponential series. The number of $r$-permutations is $$ r![x^r]\prod_{i=0}^{k-1} E_{n_i}(x) $$ where $[x^r]f(x)$ is the coefficient of $x^r$ in the polynomial $f(x)$.
On a side note, I disagree with the formula $n!/\prod n_i!$ for the number of $r$-permutations where each object appears at least once (each $t_i>0$). First, this does not involve $r$ at all. Second, in the case where the multiset is $\{A,A,B,B\}$ and $r=2$, the answer should be two, since the valid permutations are $AB$ and $BA$, but your formula gives $4!/(2!\cdot 2!)=6$. Instead, $n!/\prod n_i!$ gives the number of $n$-permutations of a multiset with $n$ elements total (all objects used completely, each $t_i=n_i$).