I can improve a little on MJD's method. This was based on computing for each state $S$ (the last $N$ values) the expected remaining number of steps $L(S)$ until a final state (last $N$ values equal) is reached.
Let me change the notation slightly. Let $T_S$ be the number of steps from state $S$ until a final state is reached, and assume that this final state consists of $N$ copies of the value $F_S$. Both $T_S$ and $F_S$ are random variables, and $L(S)=\text{E}[T_S]$.
We can then write $ \text{E}[T_S] =\sum_{u\in\mathcal{A}} \text{E}[T_S|F_S=u]\cdot\Pr[F_S=u] =\sum_{u\in\mathcal{A}}\sum_{t=0}^\infty \Pr[T_S=t, F_S=u]\cdot t $ where $u$ runs through the values of $\mathcal{A}=\{1,\ldots,N\}$. The probability $\Pr[T_S=t, F_S=u]$ is that of reaching a final state with the final $N$ values all equal to $u$ after $t$ steps.
Instead of computing the expected remaining steps for of all combinations $\mathcal{A}^N$ (modulo permutations on $\mathcal{A}$), all we need are values $ L_u(S)=\text{E}[T_S|F_S=u]\cdot\Pr[F_S=u] $ which only depends on which elements of $S$ are equal to $u$. Thus, it suffices to compute $L_1(S)$ for $S\in\{1,0\}^N$ where $1$ indicates the value $u$ and $0$ a value other than $u$.
This reduces the computational burden to having $2^N$ different $L_1(S)$ to compute.
Update: Here are the equations required for solving $L_1(S)$. As noted by MJD in the comment, these are slightly different.
For ease of notation, let's restrict ourselves to the case where $S\in\{0,1\}^N$. Then we can write $ L_1(S)=\text{E}[T_S|F_S=1]\cdot\Pr[F_S=1]=\text{E}[T_S F_S]. $ We can then express $L_1(S)$ for non-final states $S$ (i.e. not all 0s or 1s) in terms of $L_1(S')$ where $S'$ are the possible next states as $ \begin{split} L_1(S)&=\text{E}[T_S F_S]=\Pr[F_S=1]+\text{E}[(T_S-1) F_S]\\ &=\Pr[F_S=1]+\sum_{S'} \Pr[S\rightarrow S']\cdot\text{E}[T_{S'} F_{S'}]\\ &=\Pr[F_S=1]+\sum_{S'} \Pr[S\rightarrow S']\cdot L_1(S') \end{split} $ where $\Pr[S\rightarrow S']$ denotes the transition probability from $S$ to the next state $S'$: i.e. if $S=(s_1,\ldots,s_N)$ the next state will be $S'=(s_2,\ldots,s_N,1)$ with likelihood $\frac{1}{N}\sum_{i=1}^N s_i$, and $S'=(s_2,\ldots,s_N,0)$ with likelihood $1-\frac{1}{N}\sum_{i=1}^N s_i$.
Note that we could not have done this on the conditional $\text{E}[T_S|F_S=1]$, although it looks tempting, as $\text{E}[T_{S'}|F_{S'}=1]$ are conditional on $F_{S'}=1$ not $F_S=1$. However, we do have $ \Pr[F_S=1]=\sum_{S'}\Pr[S\rightarrow S']\cdot\Pr[F_{S'}=1]. $
We now need to compute $\Pr[F_s=1]$ for all $S$.
To do this, let's go back to the original $S=(1,\ldots,N)$ and let $q_k=\Pr[F_S=k]$. The next state $S'=(2,\ldots,N,s')$ for $s'=1,\ldots,N$, each with likelihood $1/N$. This gives us, writing $q_0=0$ for convenience, $ q_k=q_{k-1}+\frac{q_N}{N} \implies q_k=\frac{2k}{N(N+1)} $ which is just the above expression for $\Pr[F_S=k]$ in terms of the sum over $S'$.
For $S\in\{0,1\}^N$ we then get $\Pr[F_S=1]=\sum_{k=1}^N s_kq_k=\frac{2}{N(N+1)}\sum_{k=1}^N s_k.$
Example: Let's do $N=2$.
For ease of notation, I'll write $f_S=\Pr[F_S=1]$ and $l_S=L_1(S)$ for $S\in\{0,1\}^N$.
First $f_{00}=0$, $f_{10}=1/3$, $f_{01}=2/3$ and $f_{11}=1$.
For the final states, we have $l_{00}=l_{11}=0$. Then, we have $ \begin{split} l_{10}&=f_{10}+\frac{1}{2}(l_{00}+l_{01})=\frac{1}{3}+\frac{1}{2}l_{01}\\ l_{01}&=f_{01}+\frac{1}{2}(l_{10}+l_{11})=\frac{2}{3}+\frac{1}{2}l_{10}\\ \end{split} $ which solves to $l_{01}=10/9$ and $l_{10}=8/9$.
Going back to the original problem, $S\in\{1,\ldots,N\}$, we compute $ L(12)=\sum_u L_u(12)=l_{10}+l_{01}=2 $ where the two $l$ terms correspond to picking $u=1$ and $u=2$.
The notation could perhaps been less confusing if I'd used values true and false instead of 1 and 0, intruduced the indicator map $\chi_u(S)$ which maps $u$ to true and the other values to false, and consistenty written $S\in\mathcal{A}^N$ while using $\chi_u(S)\in\{\textit{false},\textit{true}\}^N$. However, I hope it was still clear enough.
I did the computations (using Maple to solve the linear equations). If $L_N=L(12\ldots N)$, I found: $ \begin{split} L_2&=2\\ L_3&=\frac{19}{4}=4.75\\ L_4&=\frac{1179}{140}=8.4214\ldots\\ L_5&=\frac{12226997}{934830}=13.079\ldots\\ L_6&=\frac{633096670030808}{33784478422065}=18.739\ldots\\ &\text{etc.}\\ \end{split} $ which are not numbers that factor nicely.