Because of the special choice of numbers, we can find a simple expression for the probability.
Call a number $\ge 4$ a success. The probability of success when tossing a single die is $\frac{3}{6}$, that is, $\frac{1}{2}$.
By symmetry, the probability of $3$ or more successes in $5$ trials is the same as the probability of $3$ or more failures. But the events "$3$ or more successes" and "$3$ or more failures" are disjoint, and together they encompass all possibilities. It follows that the required probability is $\frac{1}{2}$.
Remark: For the general problem, there will not be a simple expression. You can easily determine the probability $p$ of success on any one trial. The probability of $k$ successes in $n$ trials is $\binom{n}{k}p^k(1-p)^{n-k}$. Add up these expressions over the desired range. So in the case of your parameters, we have $n=C$, and we sum from $k=A$ to $k=C$.
Representation via generating functions
This isn't satisfactory in the sense that we still cannot obtain a closed form, but the representation is concise and easily programmable. Suppose we have $(k_6, k_8, k_{10}, k_{12})$ dice of types d6, d8, d10, and d12 respectively. Let
\begin{align*}
f_6(x) &= \left(\frac{5}{6}+\frac{1}{6}x\right)^{k_6} \\
f_{8}(x) &= \left(\frac{5}{8}+\frac{2}{8}x+\frac{1}{8}x^2\right)^{k_8} \\
f_{10}(x) &= \left(\frac{5}{10}+\frac{2}{10}x+\frac{2}{10}x^2+\frac{1}{10}x^3\right)^{k_{10}}\\
f_{12}(x) &= \left(\frac{5}{12}+\frac{2}{12}x+\frac{2}{12}x^2+\frac{2}{12}x^3+\frac{1}{12}x^4\right)^{k_{12}} \\
f(x) &= f_6(x)f_8(x)f_{10}(x)f_{12}(x)
\end{align*}
Let $N$ be the random variable denoting the total number of successes (slightly different notation from your post, where you let $N$ represent the value of interest). Then, the probability of getting exactly $n$ successes is
\begin{align*}
P(N = n) =[x^n]f(x)
\end{align*}
where $[x^n]f(x)$ is the coefficient of $x^n$ of $f(x)$. The cumulative distribution function (i.e. the probability of getting $n$ successes or fewer) is
\begin{align*}
P(N \le n) = [x^n]\frac{f(x)}{1-x}
\end{align*}
And so
\begin{align*}
P(N \ge n) = 1 - [x^{n-1}]\frac{f(x)}{1-x}
\end{align*}
Finite-Sample Upper Bound
Let
\begin{align*}
K = k_6 + k_{8} + k_{10} + k_{12}
\end{align*}
and so the proportion of the $K$ dice which are d6, d8, d10, and d12 are respectively
\begin{align*}
(p_6, p_8, p_{10}, p_{12}) = (k_6, k_8, k_{10}, k_{12})/K
\end{align*}
Let $N_k \in \{0, \cdots, 4\}$ ($k = 1, \cdots, K$) be the random variable denoting the success number for each die, and
\begin{align*}
X_m = \sum_{k=1}^{K}\mathbb{I}(N_k = m)
\end{align*}
denote the number of successes produced from the $K$ dice. Then the proportion of the $K$ dice falling in each $m$ ($m = 0, \cdots, 4$), is
\begin{align*}
q_0 &= \frac{5}{6}p_6 + \frac{5}{8}p_8 + \frac{5}{10}p_{10} + \frac{5}{12}p_{12} \\
q_1 &= \frac{1}{6}p_6 + \frac{2}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_2 &= \frac{0}{6}p_6 + \frac{1}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_3 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{1}{10}p_{10} + \frac{2}{12}p_{12} \\
q_4 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{0}{10}p_{10} + \frac{1}{12}p_{12}
\end{align*}
So, $(X_0, \cdots, X_4) \sim \text{Multinomial}(K, (q_0, \cdots, q_4))$.
Finally,
\begin{align*}
P(N \ge n) &= P\left(\sum_{m=0}^{4} mX_m \ge n\right) \\
&= P\left(\exp\left(t\sum_{m=0}^{4} mX_m\right) \ge \exp(tn)\right) & z \mapsto e^{tz} \text{ is increasing for } t>0\\
&\le \frac{E\left[\exp\left(t\sum_{m=0}^{4} mX_m\right)\right]}{e^{tn}} & \text{Markov's inequality} \\
&= e^{-nt}\left(\sum_{m=0}^{4}q_m e^{mt}\right)^K \\
&= \left(\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
and so we can form the Chernoff bounds
\begin{align*}
P(N \ge n) \le \left(\inf_{t>0}\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
Example
Let's suppose we have $(k_6, k_8, k_{10}, k_{12}) = (5, 7, 11, 13)$ and want to find $P(N \ge 30)$. Then
\begin{align*}
P(N \ge 30) = 1 - [x^{29}]\frac{f(x)}{1-x} = 1- \frac{56649270689104302470179125877}{148888471031133469396697088000} \approx 0.6195
\end{align*}
Using the Chernoff bound with
\begin{align*}
K = 36, \mathbf{q} = (0.5405, 0.1931, 0.1456, 0.0907, 0.0301)
\end{align*}
We find that the infimum is attained at $t^* = 0.0894$ giving us $P(N \ge 30) \le 0.8453$.
Best Answer
Using the notation $A^{k}$ to mean $k$ $A$'s:
$$\begin{eqnarray} |A^{1+} \cap B^{2+}| &=& |A^{1+}| + |B^{2+}| - |A^{1+}\cup B^{2+}| \\ &=& 3^n - |A^0| + 3^n - |B^0| - |B^1| - (3^n - |A^0\cap B^0| - |A^0\cap B^1|) \\ &=& 3^n - 2^n + 3^n - 2^n - n2^{n-1} - (3^n - 1 - n) \\ &=& 3^n - 2^{n+1} - n2^{n-1} + n + 1 \end{eqnarray}$$
So your probability is:
$$P(n) = \frac{3^n - 2^{n+1} - n2^{n-1} + n + 1}{3^n}$$