By chance I noticed that $e^{-1/e}$ and $\ln(2)$ both have decimal representation $0.69\!\ldots$, and it got me wondering how one could possibly determine which was larger without using a calculator.
For example, how might Euler have determined that these numbers are different?
Here's a method that only uses rational numbers as bounds, keeping to small denominators where possible. It is elementary, but finding sufficiently good bounds requires some experimentation or at least luck.
We instead establish the inequality $$-\log \log 2 < \frac{1}{e}$$ of positive numbers, which we can see to be equivalent by taking the logarithm of and then negating both sides. Our strategy for establishing this latter inequality (which is equivalent to yet another inequality suggested in the comments) is to use power series to produce an upper bound for the less hand side and a larger lower bound for the right-hand side. To estimate logarithms, we use the identity $$\log x = 2 \operatorname{artanh} \left(\frac{x - 1}{x + 1}\right) ,$$ which yields faster-converging series and so lets us use fewer terms for our estimate.
Firstly, $$\log 2 = 2 \operatorname{artanh} \frac{1}{3} .$$ Then, since the power series $$\operatorname{artanh} u = u + \frac{1}{3} u^3 + \frac{1}{5} u^5 + \cdots $$ has nonnegative coefficients, for $0 < u < 1$ any truncation thereof is a lower bound for the series, and in particular $$\log 2 = 2 \operatorname{artanh} \frac{1}{3} > 2\left[\left(\frac{1}{3}\right) + \frac{1}{3} \left(\frac{1}{3}\right)^3 + \frac{1}{5} \left(\frac{1}{3}\right)^5\right] = \frac{842}{1215} .$$ We'll use the same power series to produce an upper bound for $-\log \log 2$, but since we're nominally computing by hand and want to avoid computing powers of a rational number with large numerator and denominator, we'll content ourselves with a weaker rational lower bound for which computing powers is faster: Cross-multiplying shows that $$\log 2 > \frac{842}{1215} > \frac{9}{13}$$ and so $$-\log \log 2 < -\log \frac{9}{13} = 2 \operatorname{artanh} \frac{2}{11} .$$ This time, we want an upper bound for $2 \operatorname{artanh} \frac{2}{11}$. When $0 < u < 1$ we have $$\operatorname{artanh} u = \sum_{k = 0}^\infty \frac{1}{2 k + 1} u^{2 k + 1} < u + \frac{1}{3} u^3 \sum_{k = 0}^\infty u^{2k} = u + \frac{u^3}{3 (1 - u^2)},$$ and evaluating at $u = \frac{2}{11}$ gives $$2 \operatorname{artanh} \frac{2}{11} < \frac{1420}{3861} < \frac{32}{87} .$$ On the other hand, the series for $\exp(-x)$ alternates, giving the estimate $$\frac{1}{e} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} > \sum_{k=0}^7 \frac{(-1)^k}{k!} = \frac{103}{280} .$$ Combining our bounds gives the desired inequality $$-\log \log 2 < \frac{32}{87} < \frac{103}{280} < \frac{1}{e} .$$
I'm assuming (partly because he did create tables) that Euler would know the values of $e$ and $\ln(2)$ to at least 4 decimal places, and your "by hand" guy should also know those. Explicitly calculating $e^{-1/e}$, however, hast to be declared "too hard"; otherwise the problem becomes a triviality.
The following reasoning is well within Euler's knowledge base, and requires no really difficult arithmetic:
Express $e^{-1/e}$ as its Taylor series, and expand each term as a Taylor series for $e^{-n}$: $$ \begin{array}{clllllc}e^{-1/e}=&&+1 &&- \frac1{e}&+\frac1{2e^2}&-\frac1{6e^3} &+\cdots \\ e^{-1/e}=& &&-1&+1&-\frac12&+\frac16&-\cdots \\&&&+\frac12&-\frac22 & +\frac{2^2}{2!\cdot 2}&-\frac{2^3}{3!\cdot 2}&+\cdots \\&&&-\frac16&+\frac36 & -\frac{3^2}{3!\cdot 6}&+\frac{3^3}{3!\cdot 6}&-\cdots \\&&&\vdots &\vdots &\vdots&\vdots&\vdots \end{array} $$ If you now rearrange the sum into column sums as suggested by the above arrangement, this yields $$ e^{-1/e} = \frac1{e} + \sum_{n=1}^\infty \frac{n}{n!} -\frac12 \sum_{n=1}^\infty \frac{n^2}{n!}+\frac16 \sum_{n=1}^\infty \frac{n^3}{n!}-\cdots $$ or $$ e^{-1/e} = \frac1{e} + \sum_{k=1}^\infty\left( \sum_{n=1}^\infty (-1)^{n+k} \frac1{k!} \sum_{n=1}^\infty \frac{n^k}{n!}\right) $$ Now for any given moderately small integer $k$, it is not too difficult (tried it by hand in each case) to use the same sort of rearrangement tricks to sum $$ \sum_{n=1}^\infty \frac{n^k}{n!} $$ And the answers (starting at $k=1$, and including the proper sign and leading factorial) are $$ \left\{\frac1{e}, 0, -\frac1{6e}, \frac1{24e}, \frac1{60e}, -\frac1{80e}, \frac1{560e} , \frac5{4032e}\right\} $$ You have to wonder where it is safe to stop: You would like a series which rapidly decreasing terms so that you can safely stop discarding a negative term, to give an expression $E$ where you know that $e{-1/e} < E$. The raw series shown does not lend confidence, but if you group terms in pairs you get $$ \left\{\frac1{e}, 0, -\frac1{6e}, \frac1{240e}, \frac{61}{20160e}, -\frac{2257}{3628800e} \cdots \right\} $$ and that tells you you can confidently stop after the $\frac{61}{20160e}$ term, getting some number which is larger than $e^{-1/e}$.
When you do this you get $$ e^{-1/e} < \frac{7589}{4032 e} \approx 0.69242 $$ Finally, know $\ln(2)$ to four decimal places, and can do the comparison.
An interesting side relation the above "proves" is that $$ e^{1-\frac1{e}} < \frac{7589}{4032} $$
