Ok, so my answer works but requires you to:
- Be insanely good at arithmetic (so maybe does not meet your criteria of doable by hand)
- Know that $\ln 10<\color{red}{2.3026}$ (this might be known to you if you frequently change bases between logarithms, or may be calculated by series expansion)
Edit (D S) :
I noticed that getting a better approximation for $\ln 10$ significantly reduces the work required for $\ln(\phi)$. So here it goes:
$$\ln(10) = \ln(2)+\ln(5) = 3\ln(2) + \ln(5/4)$$
$$\begin{align}\ln(2) &= \ln((1+1/3)/(1-1/3)) \\
&= 2\sum_{k=0}^\infty \frac{(1/3)^{2k+1}}{2k+1}\\
&< 2\left(\sum_{k=0}^3 \frac{(1/3)^{2k+1}}{2k+1}\right)+ 2\left(\sum_{k=4}^\infty \frac{(1/3)^{2k+1}}{9}\right) \\
&= 2\left(\sum_{k=0}^3 \frac{(1/3)^{2k+1}}{2k+1}\right)+\frac{(1/3)^9}{4} \\
&= \frac{1910051}{2755620} < 0.693148
\end{align}$$and
$$\begin{align}\ln(5/4) &= \ln((1+1/9)/(1-1/9)) \\
&= 2\sum_{k=0}^\infty \frac{(1/9)^{2k+1}}{2k+1}\\
&< 2\left(\sum_{k=0}^2 \frac{(1/9)^{2k+1}}{2k+1}\right)+ 2\left(\sum_{k=3}^\infty \frac{(1/9)^{2k+1}}{7}\right) \\
&= 2\left(\sum_{k=0}^2 \frac{(1/9)^{2k+1}}{2k+1}\right)+ \frac{(1/9)^5}{280} \\
&= \frac{3689393}{16533720}<0.223144
\end{align}$$
Hence $\ln10 < 3\times 0.693148+0.223144 = \color{blue}{2.302588}$.
Let $x$ be such that
$$x^{x^2} = 1000\phi$$
or$$x^2\ln(x) = 3\ln(10)+\ln(\phi)$$
We are interested in showing $$3\ln(10)+\ln(\phi)<e^2 \iff 3\ln(10)+\ln(\phi)- e^2<0$$since $x^2\ln(x)$ is increasing in $x$.
First, we approximate $\ln(\phi)$. Note that the series expansion of $\ln(1+x)$ for $|x|>1$ is given by
$$\ln(1+x) = \ln(x) - \sum_{k=1}^\infty(-1)^k\frac{x^{-k}}{k}$$So that
$$2\ln(\phi) = \ln(\phi^2) = \ln(1+\phi) = \ln(\phi) + \sum_{k=1}^\infty(-1)^{k+1}\frac{\phi^{-k}}{k}$$
$$\iff \ln(\phi) = \sum_{k=1}^\infty(-1)^{k+1}\frac{\phi^{-k}}{k}< \sum_{k=1}^{17}(-1)^{k+1}\frac{\phi^{-k}}{k}$$since the magnitude of the terms in the sum are decreasing, and we have stopped at a positive term, so this approximation is greater.
Now, to calculate negative powers of $\phi$, you can just extend the Fibonacci sequence backwards, i.e,
$$\ldots \color{blue}{-8,5,-3,2,-1,1},0,1,1,2,3 \ldots$$
For example, $\phi^{-1} = \color{blue}1\cdot \phi \color{blue}{- 1}$, $\phi^{-2} = \color{blue}{-1}\cdot \phi +\color{blue}2$, $\phi ^{-3} = \color{blue}2\cdot \phi \color{blue}{- 3}$ and so on.
In the end, you should get
$$\sum_{k=1}^{17}(-1)^{k+1}\frac{\phi^{-k}}{k} = \frac{3433663565\sqrt 5 - 7666113149}{24504480}<\color{red}{0.48123}$$
Edit:
With our new approximation of $\ln(10)$, it is enough to take the sum till 13, i.e,
$$\ln(\phi) < \sum_{k=1}^{13}(-1)^{k+1}\frac{\phi^{-k}}{k} = \frac{4109827\sqrt 5- 9120481}{144144} < \color{blue}{0.481266}$$
Next, we will approximate $e^2$.
$$e^2 = \left(\sum_{k=0}^\infty\frac{1}{k!}\right)^2>\left(\sum_{k=0}^8\frac{1}{k!}\right)^2$$
(We don't use series expansion of $e^z$ as that requires more terms.)
$$\left(\sum_{k=0}^8\frac{1}{k!}\right)^2 = \left(\frac{109601}{40320}\right)^2>\color{red}{7.38903}$$
Finally,
$$\boxed{3\ln(10)+\ln(\phi)- e^2< 3\cdot2.3026 + 0.48123- 7.38903 = 0 \ \ }$$
Phew!
Edit:
Armed with our new bounds, we have
$$\boxed{3\ln(10)+\ln(\phi) - e^2<3\cdot 2.302588+0.481266-7.38903 = 0}$$
Update DesmosTutu :
We have a trick to evaluate $\ln 10$
$$\operatorname{arcsinh}(10)=\ln(20)+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-1)!!}{2n(2n)!!(10)^{2n}},\operatorname{arcsinh}(10)\simeq 3$$
Now use expansion of $\operatorname{arcsinh(x)}$ and Engels expansion of $\ln 2$
Source: Wikipedia
