Method
This method gives a sequence that should converge to $n$ and could feasibly be done by hand, if given sufficient time. Define $f(x)=\frac{25!}{3^x}-1$.* We will aim to find $x$, such that $f(x)=0.$ The Newton-Raphson method for finding the roots of $f$ is $x_{k+1}=x_k-\frac{3^{x_k}}{25!\ln3}+\frac1{\ln3}$. We can use the approximations $\ln(3)\approx0.910$ and $3^x=e^{x\ln3}\approx e^{1.10x}$ and $\frac{1}{25!\ln3}\approx e^{-58.1}$,** to arrive at the following sequence:
$$x_{k+1}=x_k-e^{1.10x_k-58.1}+0.910$$
The figure below shows $y=x_k$ for $k$ up to $20$, for different starting values: $x_0=10,20,30,40$. The limit of the sequence should be $y\approx 52$, which is the $n$ we were looking for.

Analysis
The sequence, seems to approach $y\approx52.732$ when $x_0<52$. However, when $x_0>52$, it seems to diverge.
Whether it's easier to compute $25!$ or $e^{1.10x_k-58.1}$ by hand is hard to say, since you'd have to do long computations either way. However, for large $k$, $x_{k+1}=x_k$. This means that $e^{1.10x_k-58.1}=0.910\approx e^0$ so we can use the Maclaurin series of $e^x$ to quickly compute $e^{1.10x_k-58.1}$. Another advantage the Newton-Raphson method has is that calculation errors can be mitigated by continuing in the sequence; as long as your $x_k$ is near $n$, you should, in theory, approach $n$ by finding $x_{k+1}$. This isn't the case for computing $25!=25\times24\times\ldots$, where an error could be disastrous.
But you should note that we have used approximations in generating our sequence. And, if we didn't have a calculator, we may not know if we're approaching $n$ or $n+1$ or something else; were our approximations accurate enough? We may also need the values of constants like $\ln5$ and $\ln\pi$ but, even in a world without calculators these could be in lookup tables.
* We may be tempted to define $f(x)=25!-3^x$ but the sequence this generates converges extremely slowly to $n$. This is probably due to its very steep slope near $n$ so I used a function with a shallower slope.
** This can be calculated with Stirling's Approximation as $\begin{aligned}\ln\left(\frac{1}{25!\ln3}\right)&\approx-\ln\left(\sqrt{25\cdot2\pi}\cdot\left(\frac{25}{e}\right)^{25}\right)-\ln\ln3\\&\approx25-51\ln5-\frac12\ln{2\pi}-\ln\ln3\\&\approx-58.1\end{aligned}$