This is a generalized coupon collector's problem. Two approaches you can take to this are used in the two answers at Coupon Collecting Problem: $4$ coupons with $p_1 = p_2 = \frac{1}{8}$ and $p_3 = p_4 = \frac{3}{8}$.
In the spirit of Ross's answer, you can define a state $(j,k)$ with $0\le j,k\le3$, in which you've generated $j$ $1$s and $k$ of the other numbers at least once. Then the expected number $a_{jk}$ of remaining generations satisfies the recurrence
$$
a_{jk}=1+\frac14a_{j+1,k}+\frac k4a_{jk}+\frac{3-k}4a_{j,k+1}\;,
$$
where the indices aren't incremented beyond $3$ and the initial value is $a_{33}=0$. The result is
\begin{array}{c|cccc}
j\setminus k&0&1&2&3\\\hline
0&\frac{1915}{144}&\frac{349}{27}&\frac{25}2&12\\
1&\frac{125}{12}&\frac{88}9&9&8\\
2&\frac{25}3&\frac{22}3&6&4\\
3&\frac{22}3&6&4&0\\
\end{array}
Thus the expected number of generations is $a_{00}=\frac{1915}{144}\approx13.3$.
In the spirit of my answer, we can apply inclusion–exclusion to the four conditions $A_i$ that you've generated $i$ a sufficient number of times ($3$ for $i=1$ and $1$ otherwise). Let $N_i$ denote the number of generations required until $A_i$ is fulfilled, and let $N=\max_iN_i$ denote the number of generations required until all the conditions $A_i$ are fulfilled. Then
\begin{eqnarray*}
\mathsf E[N]
&=&
\sum_{n=0}^\infty\mathsf P(N\gt n)
\\
&=&
\sum_{n=0}^\infty\mathsf P\left(\bigvee_{i\in\{1,2,3,4\}}N_i\gt n\right)
\\
&=&
\sum_{n=0}^\infty\sum_{\emptyset\ne S\subseteq\{1,2,3,4\}}(-1)^{|S|+1}\mathsf P\left(\bigwedge_{i\in S}N_i\gt n\right)\;.
\end{eqnarray*}
Now there are two cases. If $1\notin S$, we have
$$
\mathsf P\left(\bigwedge_{i\in S}N_i\gt n\right)=4^{-n}(4-|S|)^n\;.
$$
If $1\in S$, we have
$$
\mathsf P\left(\bigwedge_{i\in S}N_i\gt n\right)=\sum_{j=0}^2\binom nj4^{-n}(4-|S|)^{n-j}\;.
$$
Splitting the sum over $S$ into these two cases, we obtain
\begin{eqnarray*}
\sum_{n=0}^\infty\sum_{\emptyset\ne S\subseteq\{2,3,4\}}(-1)^{|S|+1}\mathsf P\left(\bigwedge_{i\in S}N_i\gt n\right)
&=&
\sum_{n=0}^\infty\sum_{k=1}^3(-1)^{k+1}\binom3k4^{-n}(4-k)^n
\\
&=&
\sum_{k=1}^3(-1)^{k+1}\binom3k\frac4k
\\
&=&
12-6+\frac43
\\
&=&
\frac{22}3
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{n=0}^\infty\sum_{S\subseteq\{2,3,4\}}(-1)^{|S|}\mathsf P\left(\bigwedge_{i\in S\cup\{1\}}N_i\gt n\right)
&=&
\sum_{n=0}^\infty\sum_{j=0}^2\sum_{k=0}^3(-1)^k\binom3k\binom nj4^{-n}(4-(k+1))^{n-j}
\\
&=&
4\sum_{j=0}^2\sum_{k=0}^3(-1)^k\binom3k\left(\frac1{k+1}\right)^{j+1}
\\
&=&
12-4\left(\frac32-1+\frac14+\frac34-\frac13+\frac1{16}+\frac38-\frac19+\frac1{64}\right)
\\
&=&
\frac{859}{144}\;.
\end{eqnarray*}
Together, this is
$$
\frac{22}3+\frac{859}{144}=\frac{1915}{144}\approx13.3\;,
$$
in agreement with the first result.
