Here is an argument completely different from joriki’s; it is also a complete solution.
A cycle is an odd permutation iff its length is even, so a permutation written as a product of disjoint cycles is even iff an even number of the factors are even cycles. Let $\pi=\sigma_1\dots\sigma_k$ be a permutation of $[n]$ written as a product of $k$ disjoint cycles. Then $n=|\,\sigma_1|+\dots+|\,\sigma_k|$, so the parity of $n$ is the same as the parity of the number of cycles of odd length. Thus, $\pi$ has an even number of cycles of even length iff $n$ and $k$ have the same parity, i.e., iff $(-1)^{n+k}=1$.
Let $e_n$ be the total number of cycles in even permutations of $[n]$, let $o_n$ be the total number of cycles in odd permutations of $[n]$, and let $d_n=e_n-o_n$. The total number of permutations of $[n]$ with $k$ cycles is given by $\left[n\atop k\right]$, the unsigned Stirling number of the first kind. Each of these $\left[n\atop k\right]$ permutations contributes $k$ cycles to $e_n$ if $(-1)^{n+k}=1$, and to $o_n$ if $(-1)^{n+k}=(-1)$. Thus, each contributes $(-1)^{n-k}k$ to $d_n$, and it follows that
$$d_n=\sum_k(-1)^{n+k}k\left[n\atop k\right]\;.$$
Now $(-1)^{n+k}\left[n\atop k\right]=(-1)^{n-k}\left[n\atop k\right]$ is the signed Stirling number of the first kind, for which we have the generating function $$\sum_k(-1)^{n+k}\left[n\atop k\right]x^k=x^{\underline{n}}\;.\tag{1}$$
(Here $x^{\underline{n}}=x(x-1)(x-2)\cdots(x-n+1)$ is the falling factorial, sometimes written $(x)_n$.)
Differentiate $(1)$ with respect to $x$ to obtain
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=Dx^{\underline{n}}=(x-n+1)Dx^{\underline{n-1}}+x^{\underline{n-1}}\;,$$
where the last step is simply the product rule, since $x^{\underline{n}}=x^{\underline{n-1}}(x-n+1)$. But $$Dx^{\underline{n-1}}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}\;,$$ so
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}+x^{\underline{n-1}}\;.\tag{2}$$
Now evaluate $(2)$ at $x=1$ to get $$d_n=(2-n)d_{n-1}+[n=1]\;,\tag{3}$$ where the last term is an Iverson bracket. If we set $d_0=0$, $(3)$ yields $d_1=[1=1]=1$, which by direct calculation is the correct value: the only permutation of $[1]$ is the identity, which is even and has one cycle. It’s now a trivial induction to check that $d_n=(-1)^n(n-2)!$ for all $n\ge 1$: the induction step is
$$\begin{align*}d_{n+1}&=\Big(2-(n+1)\Big)d_n\\
&=(1-n)(-1)^n(n-2)!\\
&=(-1)^{n+1}(n-1)!\;.
\end{align*}$$
Added: It occurs to me that there’s a rather easy argument that does not use generating functions. Let $\sigma$ be any $k$-cycle formed from elements of $[n]$; then $\sigma$ is a factor in $(n-k)!$ permutations of $[n]$. Moreover, exactly half of these permutations are even unless $n-k$ is $0$ or $1$. Thus, $\sigma$ contributes to $d_n$ iff $k=n$ or $k=n-1$.
There are $(n-1)!$ $n$-cycles; they are even permutations iff $n$ is odd, so they contribute $(-1)^{n-1}(n-1)!$ to $d_n$.
There are $n(n-2)!$ $(n-1)$-cycles: there are $n$ ways to choose the element of $n$ that is not part of the $(n-1)$-cycle, and the other $n-1$ elements can be arranged in $(n-2)!$ distinct $(n-1)$-cycles. The resulting permutation of $[n]$ is even iff $n-1$ is odd, i.e., iff $n$ is even, so they contribute $(-1)^nn(n-2)!$ to $d_n$.
It follows that $$\begin{align*}d_n&=(-1)^nn(n-2)!+(-1)^{n-1}(n-1)!\\
&=(-1)^n\Big(n(n-2)!-(n-1)!\Big)\\
&=(-1)^n(n-2)!\Big(n-(n-1)\Big)\\
&=(-1)^n(n-2)!\;.
\end{align*}$$
We count the complement, the permutations that have a cycle of length $k$, where $k$ ranges from $12$ to $20$. Then we add up. For smallish $k$, like $5$, this could be unpleasant, since one can have several $5$-cycles. But for the $k$ in our range, a permutation has at most one $k$-cycle.
Let's do the calculation. The idea has already been covered by amWhy, so we do it quickly.
The $k$ objects in the $k$-cycle can be chosen in $\binom{20}{k}$ ways. There are $(k-1)!$ circular permutations of $k$ objects. And there are $(20-k)!$ ways to permute the rest, for a total of
$$\binom{20}{k}(k-1)!(20-k)!.$$
This simplifies nicely to $\dfrac{20!}{k}$. So the complement of our set has size
$$20!\left(\frac{1}{12}+\frac{1}{13}+\cdots+\frac{1}{20}\right).$$
Remark: The result may be numerically surprising. The probability that a permutation has a large cycle is quite big.
Best Answer
We attempt to find a recurrence relation for the $e_n$, and convert that to information about the exponential generating function $E(x)=\sum_n e_n\frac{x^n}{n!}$.
Sort the permutations of $n$ with all cycles of even length by the length of the cycle that includes $1$. If this cycle has length $2k$, then there are $\binom{n-1}{2k-1}$ ways to choose the other $2k-1$ elements in the cycle, $(2k-1)!$ different cycles on those $2k$ elements, and $e_{n-2k}$ permutations of the $n-2k$ elements outside that cycle with all cycles of even length. Therefore, $$e_n = \sum_{k=1}^{\lfloor n/2\rfloor}\binom{n-1}{2k-1}\cdot (2k-1)!\cdot e_{n-2k} = \sum_{k=1}^{\lfloor n/2\rfloor}\frac{(n-1)!}{(n-2k)!}\cdot e_{n-2k}$$ $$n\frac{e_n}{n!} =\sum_{k=1}^{\lfloor n/2\rfloor}\frac{e_{n-2k}}{(n-2k)!}$$ Now, multiply by powers of $x$ to convert that to a generating function. We have that $xE'(x)=\sum_n ne_n\frac{x^n\cdot x}{n!}$, so \begin{align*}n\frac{e_n}{n!}x^n &= \sum_{k=1}^{\lfloor n/2\rfloor}\frac{e_{n-2k}x^{n-2k}}{(n-2k)!}\cdot x^{2k}\\ xE'(x) &= E(x)\cdot (x^2+x^4+\cdots) = \frac{x^2}{1-x^2}E(x)\\ \frac{E'(x)}{E(x)} &= \frac{x}{1-x^2}\\ \ln(E(x)) &= \int \frac{x}{1-x^2}\,dx = C-\frac12\ln(1-x^2)\\ E(x) &= A(1-x^2)^{-\frac12}\end{align*} We can lock down the constant of integration by evaluating at $x=0$; $e_0=1$, so $E(0)=1$ and $A=1$. There it is.