Closed form for $\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2$? 
Is there a closed form for this series:
$$\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2 \approx 1.273278374727530507449$$

(Mathematica computation by Patrick Stevens).
This is basically a sum of squared errors for all $n$ for the classic limit used to define $e$.

Are there some other similar series of interest, representing a sum of squared errors? (I'm aware we can buid an infinite set of such series by using various limits for various constants, but I'm asking only about well known series).

I don't really have motivation except for the fact that this series seems fundamental enough to have been studied before.
Besides, there exists a special value for an infinite product:
$$\prod_{k=2}^{\infty} e \left(1-\frac{1}{k^2} \right)^{k^2}=\frac{\pi}{e^{3/2}}$$
(The link had been here, but it's broken now).

Update
Some attempts to rearrange the series:
$$e=\sum_{k=0}^\infty \frac{1}{k!}$$
$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n \left( \begin{array}( n \\ k \end{array} \right) \frac{1}{n^k}$$
Thus, we can write the general term as:
$$\left(e-\left(1+\frac{1}{n}\right)^n \right)^2=\left( \sum_{k=0}^n \frac{1}{k!} \left(1-\frac{n!}{(n-k)!} \frac{1}{n^k} \right)+ \sum_{k=n-1}^\infty \frac{1}{k!} \right)^2$$
 A: This is not an answer.
I have been a bit surprised by the result given by Wolfram Alpha. The first result I got from it was $1.26411$ and asking for more digits $1.2686765$ which corresponds to what you wrote.
I computed the partial sums
$$S_p=\sum_{n=1}^{10^p}\left(e-\left(1+\frac{1}{n}\right)^n \right)^2 $$ and got the following numbers
$$\left(
\begin{array}{cc}
 p & S_p \\
 1 & 1.111547861 \\
 2 & 1.255063881 \\
 3 & 1.271433724 \\
 4 & 1.273093674
\end{array}
\right)$$
A: Only some hints.
Be $\,W(x)\,$ the main branch of the Lambert W-function .
Based on my answer for Value of the series $\sum_\limits{n=1}^{\infty}\left(\frac{n+1}{n\cdot 2\pi}\right)^ n$ we get: 
$\displaystyle f(x):=\sum\limits_{n=1}^\infty x^{n-1}\left(1+\frac{1}{n}\right)^n=-\int\limits_0^\infty \left(\frac{d}{dx}\left(\frac{W(-xte^{-t})}{xte^{-t}(1+W(-xte^{-t}))}\right)\right) dt$
$\displaystyle g(x):=\sum\limits_{n=1}^\infty x^{n-1}\left(1+\frac{2}{n}\right)^n=\int\limits_0^\infty \left(\frac{d}{dx}\left(\frac{W(-xte^{-t})^2}{(xte^{-t})^2(1+W(-xte^{-t}))}\right)\right) dt$ 
We have $\enspace \displaystyle \sum\limits_{n=1}^\infty x^{2n-2}\left(e-\left(1+\frac{1}{n}\right)^n\right)^2 = \frac{e^2}{1-x^2} – 2ef(x^2)+\frac{g(x)-g(-x)}{2x} $ 
and therefore $\enspace\displaystyle \sum\limits_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n\right)^2 = \lim\limits_{x\uparrow 1} \left(\frac{e^2}{1-x^2} – 2ef(x^2)+\frac{g(x)-g(-x)}{2x}\right) \,$ .  
Note: $\enspace$ Please understand that I don't have fun to calculate this. :-)
