It can probably be done by looking at the sum of squares of sizes of color clusters and then constructing an appropriate martingale. But here's a somewhat elegant solution: reverse the time!

Let's formulate the question like that. Let $F$ be the set of functions from $\{1,\ldots,n\}$ to $\{1,\ldots,n\}$ that are almost identity, i.e., $f(i)=i$ except for a single value $j$. Then if $f_t$ is a sequence of i.i.d. uniformly from $F$, and
$$g_t=f_1 \circ f_2 \circ \ldots \circ f_t$$
then you can define $\tau= \min \{ t | g_t \verb"is constant"\}$. The question is then to calculate $\mathbb{E}(\tau)$.

Now, one can also define the sequence
$$h_t=f_t \circ f_{t-1} \circ \ldots \circ f_1$$
That is, the difference is that while $g_{t+1}=g_t \circ f_{t+1}$, here we have $h_{t+1}=f_{t+1} \circ h_t$. This is the time reversal of the original process.

Obviously, $h_t$ and $g_t$ have the same distribution so
$$\mathbb{P}(h_t \verb"is constant")=\mathbb{P}(g_t \verb"is constant")$$
and so if we define $\sigma=\min \{ t | h_t \verb"is constant"\}$ then $\sigma$ and $\tau$ have the same distribution and in particular the same expectation.

Now calculating the expectation of $\sigma$ is straightforward: if the range of $h_t$ has $k$ distinct values, then with probability $k(k-1)/n(n-1)$ this number decreases by 1 and otherwise it stays the same. Hence $\sigma$ is the sum of geometric distributions with parameters $k(k-1)/n(n-1)$ and its expectation is
$$\mathbb{E}(\sigma)=\sum_{k=2}^n \frac{n(n-1)}{k(k-1)}= n(n-1)\sum_{k=2}^n \frac1{k-1} - \frac1{k} = n(n-1)(1-\frac1{n}) = (n-1)^2 .$$