Here's a way to do it:
(I may be writing my state vectors and transition matrices transposed relative to the way you might have learned them, or even the way they're usually done. If that's the case you'll need to translate back.)
The probability model gives you probabilities for 4 output states at time $t$ in terms of the 16 input states - the possible ordered pairs for $(x_{t-1},x_{t-2})$.
For speed of writing, let's write $AC$ for $(A,C)$ and so on.
\begin{array}{c|cccc|cccc|c}
& AA & AC &AT&AG& CA &CC &CT &CG& \ldots \\ \hline
A &p_{AA\to A}&p_{AC\to A}&p_{AT\to A}&p_{AG\to A}&p_{CA\to A}&p_{CC\to A}&p_{CT\to A}&p_{CG\to A}\\
C &p_{AA\to C}&p_{AC\to C}&p_{AT\to C}&p_{AG\to C}&p_{CA\to C}&p_{CC\to C}&p_{CT\to C}&p_{CG\to C}\\
T &p_{AA\to T}&p_{AC\to T}&p_{AT\to T}&p_{AG\to T}&p_{CA\to T}&p_{CC\to T}&p_{CT\to T}&p_{CG\to T}\\
G &p_{AA\to G}&p_{AC\to G}&p_{AT\to G}&p_{AG\to G}&p_{CA\to G}&p_{CC\to G}&p_{CT\to G}&p_{CG\to G}\\ \hline
\end{array}
We could label the partitions with boldface versions of the state $x_{t-1}$:
\begin{array}{c|cccc|cc|c|cc}
& AA & AC &AT&AG& CA & &\ldots && \ldots& GG \\ \hline
A & & & & & & & & & & \\
C & & \mathbf{A} & & & \quad\mathbf{C} & &\mathbf{T}& & \mathbf{G} & \\
T & & & & & & & & & & \\
G & & & & & & & & & & \\ \hline
\end{array}
As I mentioned in comments, you need to extend the state. Let $z_t$ consist of pairs of states $(x_t,x_{t-1})$ and now consider a Markov Chain in $z_t$; that is, you have a transition matrix $p(z_t|z_{t-1})$.
So the state at time $t$ will be one of the 16 pairs $(A,A), (A,C) \ldots (G,G)$, and the transition matrix will be a 16 $\times$ 16 matrix of transition probabilities that will be mostly zero (necessarily so, because any pair that doesn't have the second component of $z_t$ match with the first component of $z_{t-1}$ is impossible).
As above, for speed of writing, let's also write $AC$ for $(A,C)$ and so on.
For ease of display I am going to define $z_{t-1}^*$ which is simply a permuted $z_{t-1}$. We can write $p(z_t|z_{t-1}^*)$ and then arrive back at $p(z_t|z_{t-1})$ by simple permutation.
So the transition matrix for $p(z_t|z_{t-1}^*)$ is of the form:
\begin{array}{c|cccc|cc|c|cc}
& AA & AC &AT&AG& CA & &\ldots && \ldots& GG \\ \hline
AA & & & & & & & & & & \\
CA & & \mathbf{A} & & & \quad\mathbf{0} & &\mathbf{0}& & \mathbf{0} & \\
TA & & & & & & & & & & \\
GA & & & & & & & & & & \\ \hline
AC & & & & & & & & & & \\
\vdots & & \mathbf{0} & & & \quad\mathbf{C} & &\mathbf{0}& &\mathbf{0} & \\
\vdots & & & & & & & & & & \\\hline
\vdots & & \mathbf{0} & & &\quad\mathbf{0} & &\mathbf{T}& & \mathbf{0} &\\
\vdots & & & & & & & & & & \\ \hline
\vdots & & \mathbf{0} & & &\quad\mathbf{0} & &\mathbf{0}& & \mathbf{G} & \\
GG & & & & & & & & & & \\
\hline
\end{array}
We can then rearrange either the rows or columns so they're in the same order; the transition matrix no longer has that simple structure, but contains the same values.
Generally, you can use this procedure to transform any $k$-th order Markov chain to a first-order MC (also holds for Hidden Markov Models).
Best Answer
Without any actual events or a physical model that indicates the transition, I think the best you can do is a reasonable upper bound on the probability. The probability could, in fact, be zero.
Assuming N opportunities to make the transition, 0 of which made that particular one, then we could assume that the probability of what we saw is has no less than a 50% probability. In this case:
$P[0$ of $N] \ge 0.5, (1-p)^N \ge 0.5, 1-p \ge 0.5^{\frac{1}{N}}$,
$p \le 1 - 0.5^{\frac{1}{N}}$
Note that this upper bound gets small very quickly with large $N$.
For an even greater confidence, one could say that the probability of the observed event is no less than 5% probable, thus:
$p \le 1 - 0.05^{\frac{1}{N}}$
Here, alot more data is needed to get a low upper bound, but you can have pretty high confidence in it.
I'm guessing what you'd really like is a lower bound other than 0, but I'm afraid we can't get there from here.