For the symmetric case when $p_i=1/d$ for every $i$, see here. The answer is
$$
E(T)=\sum_{k=1}^d\frac{d}{k},
$$
where $T$ is the number of times the die has been rolled at the first instant when all the possible outcomes have been sampled at least once.

In the general case, the event $[T\ge n+1]$ means that at least one outcome has not been sampled yet, hence $[T\ge n+1]$ is the union over $k$ of the events $A_n(k)$ where $A_n(k)=[$The outcome $k$ has not been sampled during the first $n$ rolls$]$.

For every subset $I$ of $\{1,2,\ldots,d\}$, let $A_n(I)$ denote the intersection of the events $A_n(k)$ for $k$ in $I$. By inclusion-exclusion $P(T\ge n+1)$ is the sum of $(-1)^{|I|-1}P(A_n(I))$ over every non empty subset $I$ of $\{1,2,\ldots,d\}$.

For every $I$, $A_n(I)$ means that the $n$ first outcomes have been chosen in the complement of $I$, hence $P(A_n(I))=(1-p(I))^n$ where $p(I)$ is the sum of $p_k$ over $k$ in $I$. Hence,
$$
E(T)=\sum_{n\ge0}P(T\ge n+1)=\sum_{I\ne\emptyset}(-1)^{|I|-1}\sum_{n\ge0}(1-p(I))^n=\sum_{I\ne\emptyset}(-1)^{|I|-1}\frac1{p(I)}.
$$
This can be rewritten as
$$
E(T)=\sum_k\frac1{p_k}-\sum_{k_1\ne k_2}\frac1{p_{k_1}+p_{k_2}}+\sum_{k_i\ne k_j}\frac1{p_{k_1}+p_{k_2}+p_{k_3}}+\cdots+(-1)^{d-1}.
$$
It is an interesting exercise to recover the formula of the symmetric case from this one. (Hint: generating functions.)