To have failed means that: At least two $1$'s have occurred before the first $2$ or At least two $1$'s have occurred before the first $3$ or At least two $1$'s have occurred before the first $4$ etc...
We continue with inclusion-exclusion over these events.
So, what is the probability that at least two $1$'s have occurred before the first $2$? Well, we can conveniently forget about all other possibilities of the dice and all rolls which resulted in a result other than these. At least two $1$'s before the first $2$ when considering only $1$'s and $2$'s as possibilities occurs when the first two results which were $1$'s or $2$'s were both $1$'s and occurs with probability $\frac{1}{4}$. Similarly so had it been at least two $1$'s before the first $3$ and so on...
What is the probability that at least two $1$'s have occurred before the first $2$ and before the first $3$? Well, again, the first two rolls will both need to be $1$'s (again, remembering that we are considering the only possible rolls as being $1$'s, $2$'s, and $3$'s at the moment) and occurs with probability $\frac{1}{9}$. Similarly for other combinations of two numbers.
The pattern continues, the probability of at least two $1$'s occurring before any of a group of $k$ other results occurs with probability $\frac{1}{(k+1)^2}$ as again the first two rolls must both be $1$'s.
Remembering how many choices there are for each collection of other results of various sizes, we are now ready to put all of this together to the probability of failure as being:
$$5\times \frac{1}{4} - \binom{5}{2}\frac{1}{9}+\binom{5}{3}\frac{1}{16}-\binom{5}{4}\frac{1}{25}+\frac{1}{36} = \frac{71}{120}$$
Giving a final answer to your original question of:
$$\frac{49}{120} = 0.408\overline{3}$$
Best Answer
Let $S_k$ be all the outcomes in which a $k$ is not rolled. For each $k$, $|S_k|=(n-1)^r$ and there are $\binom{n}{1}$ choices for $k$. For each $j\ne k$, $|S_j\cap S_k|=(n-2)^r$ and there are $\binom{n}{2}$ choices for $j$ and $k$. Continuing in this manner and using Inclusion-Exclusion to count the number of outcomes missing at least $1$ number, we get $$ \begin{align} \left|\bigcup_{k=1}^nS_k\right| &=\sum_{k=1}^n|S_k|-\sum_{j< k}|S_j\cap S_k|+\sum_{i<j<k}|S_i\cap S_j\cap S_k|-\dots\\ &=\binom{n}{1}(n-1)^r-\binom{n}{2}(n-2)^r+\binom{n}{3}(n-3)^r-\dots \end{align} $$ Since there are a total of $n^r$ total outcomes, we get the number of outcomes in which all possible numbers are rolled is $$ n^r-\binom{n}{1}(n-1)^r+\binom{n}{2}(n-2)^r-\binom{n}{3}(n-3)^r+\dots $$ Thus, the probability of this occurring is $$ 1-\binom{n}{1}\left(1-\frac1n\right)^r+\binom{n}{2}\left(1-\frac2n\right)^r-\binom{n}{3}\left(1-\frac3n\right)^r+\dots $$ So we need to find the smallest $r$ so that $$ p\le\sum_{k=0}^n(-1)^k\binom{n}{k}\left(1-\frac kn\right)^r $$ Expected Duration
The expected duration until all numbers are rolled can be computed using the formulas above; i.e. $$ \begin{align} \mathrm{E}[r] &=\color{#C00000}{\sum_{r=1}^\infty}\sum_{k=0}^n(-1)^k\binom{n}{k}\color{#C00000}{r\left[\left(1-\frac kn\right)^r-\left(1-\frac kn\right)^{r-1}\right]}\\ &=\sum_{k=1}^n(-1)^k\binom{n}{k}\color{#C00000}{\left(-\frac nk\right)}\\ &=n\sum_{k=1}^n(-1)^{k-1}\color{#00A000}{\binom{n}{k}}\frac1k\\ &=n\sum_{k=1}^n(-1)^{k-1}\color{#00A000}{\sum_{j=k}^n\binom{j-1}{k-1}}\frac1k\\ &=n\sum_{j=1}^n\sum_{k=1}^j(-1)^{k-1}\binom{j-1}{k-1}\frac1k\\ &=n\sum_{j=1}^n\color{#0000FF}{\sum_{k=1}^j(-1)^{k-1}\binom{j}{k}}\frac1j\\ &=n\sum_{j=1}^n\frac1j \end{align} $$ However, there is a simpler way. Since the expected duration for a binomial event with probability $p$ is $\frac1p$, we get that the expected duration for the $k^{\mathrm{th}}$ number is $\frac{n}{n-k+1}$. Therefore, the expected duration to get all numbers is $$ \sum_{k=1}^n\frac{n}{n{-}k{+}1}=n\sum_{k=1}^n\frac1{k} $$