An equivalent way of phrasing this is that you roll 6d20 (because $6=5+2-1$ is the number of rolls after which you'll definitely have either succeeded or failed, and can't have done both), and succeed if at least two of those are higher than the threshold number. So your overall failure probability is
$$p^6 +6 p^5(1-p) \, ,$$
where $p=13/20$ is the probability of failing on any given roll (the first term counts the probability of failing 6 times, the second the probability of failing 5 times and succeeding once).
So, in your specific example, the probability of success is
$$1-(13/20)^6-6(13/20)^5(7/20) \approx 0.6809 \, .$$
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
Here is a blogpost that gives you an overview of the distributions of summed dice as the number of dice increases. In short, as the number increases, it becomes increasingly well modelled by the normal distribution.
However, there is a small gap between the analytic solution that we get for the probability distribution of dice and the normal distribution. My intuition is that this will not be important - because rolling the vast number of dice that would allow statistical tests to detect this gap is probably not something you will do in this game.
If you are happy using d6s then a roll of $n$d6 is relatively well approximated by
$ \mu = \frac{6n+1}{2}$, $ \sigma = 0.175\sqrt n $
when $n$ gets sufficiently large. I imagine I would use $n=4$ or 5.