We have :
\begin{aligned}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(-1\right)^{n+1}\left(2\pi\right)^{2n}}{\left(2n-k+1\right)!\left(2\pi\,\mathrm{i}\right)^{k}}}}&=-\frac{1}{2\pi\,\mathrm{i}}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(2\pi\,\mathrm{i}\right)^{2n-k+1}}{\left(2n-k+1\right)!}}}\\ &=-\frac{1}{2\pi\,\mathrm{i}}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(2\pi\,\mathrm{i}\right)^{k}}{k!}}}\\ &=\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi\right)^{2n}}{\left(2n\right)!}\int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}\ \ \ \ \ \left(*\right)\end{aligned}
Let $ n\in\mathbb{N}^{*} $, we have the following :
\begin{aligned}\small\left\vert\sum_{k=1}^{n}{\left(-1\right)^{k}\frac{\left(2\pi\right)^{2k}}{\left(2k\right)!}\int_{0}^{1}{x^{2k}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}-\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x}\right\vert &\small =\left\vert\int_{0}^{1}{\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{\left(2\pi x\right)^{2k}}{\left(2k\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x}\right\vert\\ &\small\leq\int_{0}^{1}{\left\vert\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{\left(2\pi x\right)^{2k}}{\left(2k\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\right\vert\mathrm{d}x}\\ &\small\leq\int_{0}^{1}{\frac{\left(2\pi x\right)^{2n+2}}{\left(2n+2\right)!}\,\mathrm{d}x}=\frac{\left(2\pi\right)^{2n+2}}{\left(2n+3\right)!}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Thus : $$\small \sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi\right)^{2n}}{\left(2n\right)!}\int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\left(-1\right)^{k}\frac{\left(2\pi\right)^{2k}}{\left(2k\right)!}\int_{0}^{1}{x^{2k}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}} =\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x} $$
$$ \text{(In other words we can interchange the sum with the integral)} $$
We get that : \begin{aligned}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(-1\right)^{n+1}\left(2\pi\right)^{2n}}{\left(2n-k+1\right)!\left(2\pi\,\mathrm{i}\right)^{k}}}}&=\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x} \\&=\int_{0}^{1}{\left(\cos{\left(2\pi x\right)}-1\right)\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}\\ &\small =\frac{1}{2}\end{aligned}
To get $ \left(*\right) $, we can integrate by parts and prove that : $$ \int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}=-\frac{\left(2n\right)!}{\left(2\pi\right)^{2n+1}}\sum_{k=1}^{2n}{\frac{\left(2\pi\mathrm{i}\right)^{k}}{k!}} $$
How about this:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\sum_{n=0}^{\infty}\frac{2}{\sqrt{\pi}}\int_n^{\infty}e^{-t^2}\, dt
$$
In this sum of integrals, the interval $[0,1)$ will be counted only once, in the $n = 0$ term. The interval $[1,2)$ will be counted twice, in the $n = 0$ and $n = 1$ terms. And so on. This means we can write:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\int_0^{\infty}\, \lfloor t+1\rfloor \,e^{-t^2}\, dt
$$
I don't know if this is the sort of alternative integral representation you were looking for.
Edited to add two other ways to write this expression:
First way: The quantity $\lfloor t+1\rfloor$ can be written as
$$
\lfloor t+1\rfloor \;=\; (t+1) \;-\; S(t)\, .
$$
where $S(t)$ is a sawtooth wave of period $1$ with minimum value $0$ and a maximum value $1$. One way we could write this sawtooth is as $S(t) \,=\, t\;\mathrm{mod}\;1$. Substituting this expression for $\lfloor t+1\rfloor$ into the integral above and using the fact that $\int_0^{\infty} dt \, (t+1)\,e^{-t^2} = (1+\sqrt{\pi})/2$ yields
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\left[\frac{1}{2}(1+\sqrt{\pi}) \;-\; \int_0^{\infty}dt\, S(t)\, e^{-t^2}\right].
$$
Evaluating this numerically in Mathematica with 20 digits of precision yields $1.16200283409802758182$, which is greater than Mathematica's estimate for the original sum by roughly $3.8\times {10}^{-6}$. Close enough for the vagaries of numerically evaluating weird sums and integrals.
Second way: Poisson's summation formula states that if $f(x)$ is a function and
$$
\hat{f}(q) \;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, f(x)
$$
is its Fourier transform, then
$$
\sum_{n = -\infty}^{+\infty} f(n) \;=\; \sum_{n=-\infty}^{+\infty} \hat{f}(2\pi n)\, .
$$
Define the even function $f(x) = \mathrm{erfc}(|x|)$. Since $f$ is even, we can write the original sum as
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{1}{2}f(0) \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} f(n)
\;=\; \frac{1}{2} \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} \hat{f}(2\pi n)\, .\hspace{0.5in}\text{(1)}
$$
The Fourier transform of $f$ is:
\begin{align}
\hat{f}(q) &\;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, \mathrm{erfc}(|x|)\\[0.1in]
&\;=\; 2\int_0^{+\infty} dx\, \cos(q x)\, \mathrm{erfc}(x)\hspace{0.5in}\text{Since $\mathrm{erfc}(|x|)$ is even}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dx\, \cos(q x)\,\int_x^{\infty}dt\, e^{-t^2} \hspace{0.5in}\text{Definition of $\mathrm{erfc}$}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\int_0^t dx\, \cos(q x) \hspace{0.5in}\text{Reverse order of integration}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\frac{\sin(q t)}{q}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\frac{D(q/2)}{q}
\end{align}
In the last line above, "$D$" is the Dawson function according to the third definition here.
This expression for $\hat{f}(q)$ is valid everywhere except at $q = 0$, where
$\hat{f}(0) = 2/\sqrt{\pi}$. Plugging all of this into (1) results in:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n)
\;=\; \frac{1}{2}\left[1 + \frac{2}{\sqrt{\pi}} \,+\,\frac{8}{\sqrt{\pi}}\sum_{n = 1}^{\infty} \frac{D(\pi n)}{2\pi n}\right]\, .
$$
Evaluating this numerically in Mathematica yields $1.16199904795$.
Best Answer
$$E_n(n)=\int_1^{\infty } e^{-n t} t^{-n} \, dt\tag{1}$$
$$\sum _{n=1}^{\infty } e^{-n t} t^{-n}=\frac{1}{t\ e^t-1}\tag{2}$$
$$S=\sum\limits_{n=1}^\infty E_{n}(n)=\int\limits_1^\infty\frac{1}{t\ e^t-1}\,dt=0.269292\tag{3}$$
The result in (3) above was derived using numerical integration via the Mathematica NIntegrate function. Also see Wolfram Alpha evaluation of the integral.