There is an easy solution to this in principle, but maybe one that is not very handy in practice. First, observe that $A-C$ can vary between $-n$ and $n$ so let us look at $n+\Delta_1 = n+A-C$ instead to have a nonnegative discrete random variable. Let's say its mass function is
$$ p(i) = \text{Prob}[n+A-C = i] $$
Now, we can see $A$, $B$, $C$ as the result of $n$ independent throws of a three-sided die with probabilities $\pi_A$, $\pi_B$ and $\pi_B$. The result of one throw ($n=1$) has probability generating function (pgf)
$$
\text{E}[x^Ay^Bz^C|n=1] = \pi_A x + \pi_B y + \pi_C z\,,
$$
so that because of the independence between the throws the joint pgf of $A$, $B$, $C$ is
$$
F(x,y,z) = \text{E}[x^Ay^Bz^C] = (\pi_A x + \pi_B y + \pi_C z)^n\,.
$$
Since $A+B+C=n$ we know that $n+A-C=2A+B$ with pgf
$$
z^n\text{E}[z^{A-C}] = \text{E}[z^{2A+B}] = \text{E}[(z^2)^A z^B 1^C] = F(z^2,z,1) = (\pi_A z^2 + \pi_B z + \pi_C)^n\,.
$$
In principle, this gives you the mass function $p(i)$ you need since
$$
\text{E}[z^{n+A-C}] = \sum_{i=0}^{2n} p(i) z^i\,,
$$
and $p(i)$ is identified as the coefficient in the series expansion in $z$ of the above pgf. I am sure @glen_b could tell you exactly what the numerical complexity is of doing something like that.
In general, $p(i)$ is not any of the 'standard' univariate distributions but you have some things going for you if $n$ is either very small or very large. If $n$ is small, you could expand the pgf 'by hand' so to say. If $n$ is large, then the correlation between $A$ and $C$ vanishes and their distribution becomes normal due to CLT so you can assume that $A-C$ is normal as well with $\text{Var}[A-C]=\text{Var}[A]+\text{Var}[C]$, as was suggested in the comments above. The simple estimator $(A-C)/n$ for $\pi_A-\pi_C$ will be very adequate then.
