Every face at least once

Idea. The total waiting time is a sum of independent geometric waiting times, one for each new face. Linearity of expectation does the rest.

Step 1: stages. Let $T_k$ be the number of rolls needed to see the $k$-th new face once $k-1$ faces have been seen. The total is $T=T_1+T_2+\cdots +T_6$.

Step 2: each stage. With $k-1$ faces seen, a roll shows a new face with probability $p_k=\frac{6-(k-1)}{6}$, independently of the other rolls. So $T_k$ is geometric with mean $\frac{1}{p_k}=\frac{6}{7-k}$.

Step 3: add. $$\begin{array}{rl}\mathbb{E}[T]& =\frac{6}{6}+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1}\\ & =6\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\\ & =6\cdot \frac{49}{20}=\frac{147}{10}.\end{array}$$

So it takes $14.7$ rolls on average. Most of that is the last stage: the final face alone costs $6$ rolls on average.

In general. With $n$ equally likely faces the answer is $n{H}_n\approx n\ln n$, where $H_n$ is the $n$-th harmonic number.