Numerical evidence from Wolfram seems to indicate this. Is this just a numerical coincidence, and if not, can someone explain it?
Another interesting, if unrelated numerical coincidence: Wolfram's approximation of the integral of $x^{-x}$ from $0$ to $1$ is the same as its value for the sum of $n^{-n}$ from $1$ to infinity.
Partial Answer
The Puiseux series for $x^x$ is
$$x^{-x} = \sum_{n=0}^{\infty} (-1)^n\frac{x^n \log^n(x)}{n!}$$
If we take the antiderivative we get
$$\int \bigg(\sum_{n=0}^{\infty} (-1)^n\frac{x^n \log^n(x)}{n!}\bigg)dx$$
$$= \sum_{n=0}^{\infty} \int (-1)^n\frac{x^n \log^n(x)}{n!}dx$$
$$=\sum_{n=0}^{\infty}\bigg(\frac{(-1)^n}{n!}\int x^n \log^n(x)dx\bigg)$$
We then note that
$$\int x^n \log^n(x)dx = \frac{\Gamma(n+1,-(n+1)\log(x))(-n-1)^{-n} }{(n+1)}$$
Substituting, we get that
$$\int x^{-x} dx = \sum_{n=0}^{\infty}\bigg((-1)^n\frac{\Gamma(n+1,-(n+1)\log(x))(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$
Note that if we let $x=1$ we can check this series with the Sophmore's Dream (also note that the integral is $0$ when evaluated at $0$, so all we need is the upper limit)
$$\int x^{-x} dx = \sum_{n=0}^{\infty}\bigg((-1)^n\frac{\Gamma(n+1,-(n+1)\log(x))(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$
$$= \sum_{n=0}^{\infty}\bigg((-1)^n\frac{\Gamma(n+1,0)(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$
$$= \sum_{n=0}^{\infty}\bigg((-1)^n\frac{\Gamma(n+1)(-n-1)^{-n} }{\Gamma(n+2)}\bigg) = \sum_{n=0}^{\infty}(n+1)^{-(n+1)}$$
We now let $u=n+1$ and get
$$= \sum_{u=1}^{\infty}(u)^{-u} = \text{Sophmore's Dream}$$
Now that we are convinced let's head back and take the limit. We desire to find
$$\lim_{r \to \infty} \sum_{n=0}^{\infty}\bigg((-1)^n\frac{\Gamma(n+1,-(n+1)\log(r))(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$
Honestly, I can't figure out how this limit could converge, but it must if the integral is already known. Perhaps someone more versed in limits of this type could lend a hand!
Update
I fear that I may have naively stepped outside of the interval of convergence for the series, so this approach is unlikely to work as is
