So far, I know that $1<t$ and that $$\sum_{n=0}^\infty\frac{1}{n!}\int_1^t x^n (\ln{x})^n dx = \sum_{n=2}^\infty \frac{(-1)^n}{n^n}$$ I got this by using power series but I don't know where to go from here. Using a calculator I found that $t \approx 1.19491.$
Is there a closed form for $t$?
$$\int_0^t x^x dx = 1$$ As far as I know there is no closed form for $t$.
A non-standardized closed form could be defined from the inverse of the Sphd$(1,t)$ function. But this would be purely symbolic without practical interest. https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
As already pointed out by Jorge Zuniga numerical calculus provides very accurate approximate.
$$t\simeq 1.194907008026...$$ One can find an infinity of empirical equations giving very close values of $t$. For exemple : $$t\simeq \left(\frac{1}{2\,\gamma^3\,(\gamma+e^{-1})^2}\right)^{1/6}\simeq 1.194907008020...$$ $\gamma\simeq 0.577215664901533$ is the Euler–Mascheroni constant.
The coincidence holds up to twelve digits. But this is pure coincidence as one can found a lot of coincidences thanks to some methods of experimental mathematics : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales
In addition EXAMPLES OF OTHER COINCIDENCES
First example ( coincidence holding up to ten digits) :
$$\int_0^t x^x dx = e\quad\text{from numerical calculus : }\quad t\simeq 1.9703725259...$$ $$t\simeq \left(e^e+\frac{1}{\gamma} \right)^{1/3}\left(e^{-1}+e^K\right)\simeq 1.9703725262...$$ $\gamma\simeq 0.577215664901533...$ is the Euler–Mascheroni constant.
$K\simeq 0.915965594177219...= $ is the Catalan constant.
Second example ( coincidence holding up to ten digits) :
$$\int_0^t x^x dx = \pi\quad\text{from numerical calculus : }\quad t\simeq 2.0722851702...$$ $$t\simeq \frac{e^{(K^{\,3})}}{\gamma\;(e+\pi)}\simeq 2.0722851709...$$ Those empirical formulas where obtained in a few minutes on an ordinary PC with the kind of I.S.C. software described in the paper quoted above.
Of course such councidences are for amusement only. They are meaningless.
I don't think that there is any analytical solution.
For a numerical one, you can solve by Runge-Kutta the ODE
$$\frac{dt}{dI}=t^{-t}$$ for $I\in[0,1]$.
This avoids having to combine a solver and an integrator.
For the fun of not solving an equation
Using $$x^x=\sum_{n=0}^\infty \frac {a_n}{b_n}\,(x-1)^n$$ the first $a_n$ and $b_n$ corresponding to sequences $A082525$ and $A082526$ in $OEIS$ $$I=\int_0^t x^x\,dx=\sum_{n=0}^\infty \frac {a_n}{b_n}\,\frac{(t-1)^{n+1}+(-1)^n}{n+1}$$ Rewrite $I$ as $$I=\sum_{n=0}^\infty c_n\,(t-1)^n$$ Truncate to some order $p$ and use power series reversion to get rational numbers. Converted to decimals $$\left( \begin{array}{cc} p & t_{(p)} \\ 10 & 1.1992382 \\ 20 & 1.1960254 \\ 30 & 1.1953982 \\ 40 & 1.1951801 \\ 50 & 1.1950802 \\ 60 & 1.1950264 \\ 70 & 1.1949942 \\ 80 & 1.1949735 \\ 90 & 1.1949593 \\ 100 & 1.1949492 \\ 200 & 1.1949174 \\ 300 & 1.1949116 \\ \end{array} \right)$$ which, very slowly converges to the value given by @Jorge Zuniga in comments.
Now, searching the $ISC$ for simple numbers $$t\sim \frac 6 5-\frac{1}{500}\, \sqrt{\frac{3}{2}}\,\, \left(2\sqrt[3]{18}-\sqrt{10}\right)$$ which is in an absolute error of $3.39 \times 10^{-10}$ but which does not mean anything.
For this value, the result of numerical integration is $1.000000000419\cdots$.
