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TL;DR: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

$$\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}~=~\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align}$$ where $$\begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w} +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align}$$ so that $$\begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}$$

References:

1. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; subsection 2.3.5.

2. J. Bedford, arXiv:1107.3967; p. 65 eq. (A.22). (Hat tip: Sanjana.)

TL;DR: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

$$\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr ~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}\cr ~=~~~&\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align}$$ where $$\begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w}\cr & +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align}$$ so that $$\begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}$$

References:

1. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; subsection 2.3.5.

2. J. Bedford, arXiv:1107.3967; p. 65 eq. (A.22). (Hat tip: Sanjana.)

TL;DR: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

$$\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}~=~\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align}$$ where $$\begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w} +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align}$$ so that $$\begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}$$

TL;DR: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

$$\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}~=~\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align}$$ where $$\begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w} +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align}$$ so that $$\begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}$$

References:

1. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; subsection 2.3.5.

2. J. Bedford, arXiv:1107.3967; p. 65 eq. (A.22). (Hat tip: Sanjana.)

Hint: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the saddle point approximation.

TL;DR: The extra factor $n^{-3/4}$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

$$\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}~=~\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align}$$ where $$\begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w} +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align}$$ so that $$\begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}$$