The cumulative distribution function you are seeking is
$$F_n(x)=2\int_0^{x}\exp\left({-\frac{1}{4 u^2}}\right) \,\frac{\left(8 n u^2-1\right) I_n\left(\frac{1}{4
   u^2}\right)+I_{n+1}\left(\frac{1}{4 u^2}\right)}{4 |u|^3}\, du.$$
It has the general form
$$F_n(x)=e^{-1/z}\bigl(A_n(z)I_0(1/z)+B_n(z)I_1(1/z)\bigr),$$
with $A_n(z)$ a polynomial in $z=4x^2$ of degree $\max(0,n-3)$ and $B_n(z)$ a polynomial in $z$ of degree $\max(0,n-2)$. For $n\in\{1,2,\ldots 10\}$ I find
$$\{A_n\}=\left\{1,1,3,3-8 z,48 z^2-8 z+5,-384 z^3+48 z^2-32 z+5,3840 z^4-384 z^3+336 z^2-32 z+7,-46080 z^5+3840 z^4-4224 z^3+336 z^2-80 z+7,645120 z^6-46080 z^5+61440 z^4-4224 z^3+1296 z^2-80 z+9,-10321920 z^7+645120 z^6-1013760 z^5+61440 z^4-23424 z^3+1296 z^2-160 z+9\right\},$$
$$\{B_n\}=\left\{0,2,2-4 z,16 z^2-4 z+4,-96 z^3+16 z^2-20 z+4,768 z^4-96 z^3+160 z^2-20 z+6,-7680 z^5+768 z^4-1632 z^3+160 z^2-56 z+6,92160 z^6-7680 z^5+19968 z^4-1632 z^3+736 z^2-56 z+8,-1290240 z^7+92160 z^6-284160 z^5+19968 z^4-11232 z^3+736 z^2-120 z+8,20643840 z^8-1290240 z^7+4608000 z^6-284160 z^5+192768 z^4-11232 z^3+2336 z^2-120 z+10\right\}.$$
<sub>(I checked that $\lim_{x\rightarrow\infty}F_n(x)=1$, as it should be.)</sub>

Perhaps these are known sequences of polynomials, I don't know.