If you want to generate random elements of $SU(n)$ close to the identity, it seems to me that the most logical approach would be to generate random elements of the Lie algebra $\mathfrak{su}(n)$, and then exponentiate them.
The Lie algebra $\mathfrak{su}(n)$ consists of all $n\times n$ matrices $A$ with trace $0$ satisfying $A^\dagger = -A$. In particular, the diagonal entries of $A$ must be pure imaginary and add up to $0$, and the entries below the diagonal are entirely determined by the entries above the diagonal. Given a matrix $A\in\mathfrak{su}(n)$, the exponential $\exp(tA)$ is a unitary matrix for all $t\in\mathbb{R}$.
There is a standard inner product on $\mathfrak{su}(n)$, namely the negative of the Killing form, which is invariant under the adjoint action. This is given by the formula $ \langle A, B\rangle = -2n\,\mathrm{tr}(AB) = -2n\sum_{i=1}^n\sum_{j=1}^n a_{ij} b_{ji} $ We will drop the factor of $2n$. Since $b_{ji} = -b_{ij}^*$, this can also be written $ \langle A, B\rangle = \sum_{i=1}^n\sum_{j=1}^n a_{ij} b_{ij}^* \;=\; \sum_{i=1}^n \mathrm{Im}(a_{ii})\mathrm{Im}(b_{ii}) + \sum_{i In particular, $ \langle A,A\rangle \;=\; \sum_{i=1}^n |a_{ii}|^2 + \sum_{i Presumably the right thing to do is to choose random elements $A$ of $\mathfrak{su}(n)$ having unit norm with respect to this inner product, and then compute $\exp(tA)$ for some fixed small value of $t$. To choose random elements with unit norm, you should start by finding an orthonormal basis for $\mathfrak{su}(n)$ (which should be fairy easy given the above formulas, except for the added complication that the sum of the diagonal entries is $0$, which can be solved with Gram-Schmidt), and then choose coefficients for these basis vectors in the same way that you would choose the coordinates of a random point on the unit sphere in $\mathbb{R}^{n^2-1}$.