Letting \( P_n(x) = \sinh((2n+1) \mathrm{arsinh}(x)) \) (so your \( f_{2n+1}(x) = 2 P_n(x/2) \)), I noticed
\( P_0(x) = x \)
\( P_1(x) = 4x^3 + 3x \)
\( P_2(x) = 16x^5 + 20x^3 + 5x \)
\( P_3(x) = 64x^7 + 112x^5 + 56x^3 + 7x \)
\( P_4(x) = 256x^9 + 576x^7 + 432x^5 + 120x^3 + 9x \)
...
Now look at the Chebyshev polynomials \( T_n(x) \) for odd \( n \)...
\( T_1(x) = x \)
\( T_3(x) = 4x^3 - 3x \)
\( T_5(x) = 16x^5 - 20x^3 + 5x \)
\( T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x \)
\( T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x \)
...
So \( P_n(x) = \frac{T_{2n+1}(ix)}{i} = -i T_{2n+1}(ix) \). Then,
\( f_{2n+1}(x) = 2 \sinh((2n+1) \mathrm{arsinh}(x/2)) = -2i T_{2n+1}\left(\frac{ix}{2}\right) \).
This, I suppose, is as "close as we can get to something in the 'usual' toolbox", at least if your "usual" toolbox has enough to include well-known named sequences of polynomials like the Chebyshev polynomials.
Note also that
\( T_n(x) = \cos(n \mathrm{arccos}(x)) = \cosh(n \mathrm{arcosh}(x)) \)
and, for \( x > 0 \),
\( \begin{align}
\mathrm{arcosh}(ix) &= \log(ix + \sqrt{(ix)^2 - 1})\\
&= \log(ix + \sqrt{-x^2 - 1})\\
&= \log(ix + \sqrt{-(x^2 + 1)})\\
&= \log(ix + i \sqrt{x^2 + 1}) \\
&= \log(i(x + \sqrt{x^2 + 1})\\
&= \log(i) + \log(x + \sqrt{x^2 + 1})\ (\mathrm{note\ that\ }-\pi < \arg(i) + \arg(x + \sqrt{x^2 + 1}) < \pi\ \mathrm{so\ the\ log\ identity\ is\ OK})\\
&= \log(i) + \mathrm{arsinh}(x)
\end{align}
\)
. Now
\( T_{2n+1}(ix) = \sinh((2n+1) (\log(i) + \mathrm{arsinh}(x))) = \cosh((2n+1) \log(i)) \sinh((2n+1) \mathrm{arsinh}(x)) + \sinh((2n+1) \log(i)) \cosh((2n+1) \mathrm{arsinh}(x)) \)
Taking \( \log(i) = \frac{i \pi}{2} \) and using the correspondence between hyperbolic and trigonometric functions gives \( \sinh\left((2n+1)\frac{i \pi}{2}\right) = \sinh\left(i\left(n + \frac{1}{2}\right) \pi\right) = i \sin\left(\left(n + \frac{1}{2}\right)\pi\right) = i (-1)^n \). Also, for cosh we get \( \cosh\left((2n+1)\frac{i \pi}{2}\right) = \cos\left(\left(n + \frac{1}{2}\right)\pi\right) = 0 \). Thus the result above simplifies to \( T_{2n+1}(x) = i \sinh((2n+1) \mathrm{arsinh}(x)) \) (the \( (-1)^n \) drops out due to the evenness of cosh) and so we have a formal proof of the relation to the Chebyshev polynomials we just gave.
\( P_0(x) = x \)
\( P_1(x) = 4x^3 + 3x \)
\( P_2(x) = 16x^5 + 20x^3 + 5x \)
\( P_3(x) = 64x^7 + 112x^5 + 56x^3 + 7x \)
\( P_4(x) = 256x^9 + 576x^7 + 432x^5 + 120x^3 + 9x \)
...
Now look at the Chebyshev polynomials \( T_n(x) \) for odd \( n \)...
\( T_1(x) = x \)
\( T_3(x) = 4x^3 - 3x \)
\( T_5(x) = 16x^5 - 20x^3 + 5x \)
\( T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x \)
\( T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x \)
...
So \( P_n(x) = \frac{T_{2n+1}(ix)}{i} = -i T_{2n+1}(ix) \). Then,
\( f_{2n+1}(x) = 2 \sinh((2n+1) \mathrm{arsinh}(x/2)) = -2i T_{2n+1}\left(\frac{ix}{2}\right) \).
This, I suppose, is as "close as we can get to something in the 'usual' toolbox", at least if your "usual" toolbox has enough to include well-known named sequences of polynomials like the Chebyshev polynomials.
Note also that
\( T_n(x) = \cos(n \mathrm{arccos}(x)) = \cosh(n \mathrm{arcosh}(x)) \)
and, for \( x > 0 \),
\( \begin{align}
\mathrm{arcosh}(ix) &= \log(ix + \sqrt{(ix)^2 - 1})\\
&= \log(ix + \sqrt{-x^2 - 1})\\
&= \log(ix + \sqrt{-(x^2 + 1)})\\
&= \log(ix + i \sqrt{x^2 + 1}) \\
&= \log(i(x + \sqrt{x^2 + 1})\\
&= \log(i) + \log(x + \sqrt{x^2 + 1})\ (\mathrm{note\ that\ }-\pi < \arg(i) + \arg(x + \sqrt{x^2 + 1}) < \pi\ \mathrm{so\ the\ log\ identity\ is\ OK})\\
&= \log(i) + \mathrm{arsinh}(x)
\end{align}
\)
. Now
\( T_{2n+1}(ix) = \sinh((2n+1) (\log(i) + \mathrm{arsinh}(x))) = \cosh((2n+1) \log(i)) \sinh((2n+1) \mathrm{arsinh}(x)) + \sinh((2n+1) \log(i)) \cosh((2n+1) \mathrm{arsinh}(x)) \)
Taking \( \log(i) = \frac{i \pi}{2} \) and using the correspondence between hyperbolic and trigonometric functions gives \( \sinh\left((2n+1)\frac{i \pi}{2}\right) = \sinh\left(i\left(n + \frac{1}{2}\right) \pi\right) = i \sin\left(\left(n + \frac{1}{2}\right)\pi\right) = i (-1)^n \). Also, for cosh we get \( \cosh\left((2n+1)\frac{i \pi}{2}\right) = \cos\left(\left(n + \frac{1}{2}\right)\pi\right) = 0 \). Thus the result above simplifies to \( T_{2n+1}(x) = i \sinh((2n+1) \mathrm{arsinh}(x)) \) (the \( (-1)^n \) drops out due to the evenness of cosh) and so we have a formal proof of the relation to the Chebyshev polynomials we just gave.