The density of $\left(\frac{Y_{i}}{\sqrt{\sum _{i}^{n}Y^{2}_{i}} }\right)_{i}^{n}$ for iid $Y_{i}\in N(0,1)$ can be computed by change of density formula. The answer will be rotationally symmetric.
Any tricks in computing the determinant of the Jacobian (doing it by hand gets unwieldy even at $n=3$)
$\frac{1}{\sqrt{\sum x_{i}^{2}} }\begin{pmatrix} 2x_{1}^{2}+\sum x_{i\neq 1}^{2} & x_{1}x_{2} & ... &x_{1}x_{n} \\ & & & \\ & & & \\ x_{n}x_{1}& x_{n}x_{2} & ... &2x_{n}^{2}+\sum_{\neq n} x_{i}^{2} \end{pmatrix}$
I am also curious what the diagonal form of this matrix is (which also helps with computing the determinant). This matrix is symmetric so it is diagonalizable.
I did not check the computations of the Jacobian. Anyway, the determinant you wrote is equal to $\lVert x\rVert^{-1}\det\left(\lVert x\rVert^2I_n+A \right)$, where $A$ is the matrix whose entries are given by $a_{i,j}:=x_ix_j$. The characteristic polynomial of $A$ can be computed easily, since $A$ as a rank of $1$ (its range is $\operatorname{Vect}(x)$) and its trace is $\lVert x\rVert^2$. So $$\det\left(\lambda I-A\right) = \left(\lambda -\lVert x\rVert^2\right)\lambda^{n-1}. $$ It follows that $$\det\left(\lVert x\rVert^2I_n+A \right)=(-1)^n\det\left(-\lVert x\rVert^2I-A\right)=(-1)^n\left(- 2\lVert x\rVert^2\right)\left(-\lVert x\rVert^2\right)^{n-1} =2\lVert x\rVert^{2n}.$$