First of all, let us compute the probability of success of this algorithm. If you are given the state $|\psi\rangle\langle\psi|\otimes|\psi\rangle\langle\psi|$, the SWAP test will return the state $|0\rangle$ with probability $1$, which is the probability of success of the algorithm in this case.
Let us now consider the second case. The initial state is:
$$\rho_0=\frac{1}{2^{2n}}\sum_{i,j}|0,i,j\rangle\langle0,i,j|$$
The first gate to be applied is:
$$\mathbf{H}\otimes \mathbf{I}\otimes\mathbf{I}=\frac{1}{\sqrt{2}}\sum_{a,b,x,y}(-1)^{a\cdot b}|a,x,y\rangle\langle b,x,y|.$$ The resulting state is thus given by:
$$\rho_1=\frac{1}{2}\frac{1}{2^{2n}}\sum_{a,i,j,b}|a,i,j\rangle\langle b,i,j|$$
We now apply the $\mathbf{CSWAP}$ gate, whose expression is:
$$\mathbf{CSWAP}=\sum_{x,y}|0,x,y\rangle\langle0,x,y|+\sum_{x,y}|1,x,y\rangle\langle1,y,x|$$
The resulting state is:
$$\rho_2=\frac{1}{2}\frac{1}{2^{2n}}\sum_{i,j}\left(|0,i,j\rangle\langle0,i,j|+|0,i,j\rangle\langle1,j,i|+|1,j,i\rangle\langle0,i,j|+|1,j,i\rangle\langle1,j,i|\right)$$
Finally, we apply the Hadamard gate on the first qubit once again, which results in the state:
$$\rho_3=\frac{1}{4}\frac{1}{2^{2n}}\sum_{i,j}\left(\sum_{a,b}|a,i,j\rangle\langle b,i,j|+\sum_{a,b}(-1)^b|a,i,j\rangle\langle b,j,i|+\sum_{a,b}(-1)^a|a,j,i\rangle\langle b,i,j|+\sum_{a,b}(-1)^{a\oplus b}|a,j,i\rangle\langle b,j,i|\right)$$
We're interested by the diagonal coefficients of $\rho_3$ that can be written as $|0,i,j\rangle\langle0,i,j|$. Summing them would give us the probability of measuring $|0\rangle$. This probability is thus given by:
$$\mathbb{P}[|0\rangle]=\frac{1}{4}\frac{1}{2^{2n}}\left(\sum_{i,j}1+\sum_{i}1+\sum_{i}1+\sum_{i,j}1\right)=\frac12+\frac{1}{2^{n+1}}.$$
All in all, this algorithm distinguishes these two states with probability $\frac34-\frac{1}{2^{n+2}}$.
Now, let $T$ denote the trace distance between these two states. We know that the optimal probability of disinguishing these states is given by $\frac12(1+T)$. Let $U$ be a quantum gate such that $U|0\rangle=|\psi\rangle$. $T$ is then also equal to the trace distance between $\left(U^\dagger\otimes U^\dagger\right)\left(|\psi\rangle\langle\psi|\otimes|\psi\rangle\langle\psi|\right)\left(U\otimes U\right)=|0\rangle\langle0|\otimes|0\rangle\langle0|$ and $\frac{1}{2^{2n}}\left(U^\dagger\otimes U^\dagger\right)\mathbf{I}\left(U\otimes U\right)=\frac{1}{2^{2n}}\mathbf{I}$. $T$ is then easily seen to be:
$$T=\frac12\sum_i\left|\lambda_i\right|=\frac12\left(1-\frac{1}{2^{2n}}+\sum_{i=1}^{2^{2n}-1}\frac{1}{2^{2n}}\right)=1-\frac{1}{2^{2n}}$$
which means that the maximal probability of distinguishing these states is $1-\frac{1}{2^{2n+1}}$.
Thus, the SWAP test has a sub-optimal probability of success. Intuitively, this is due to the fact that the probability of measuring $|0\rangle$ is always larger than or equal to $\frac12$, which upper-bounds the probability of success with $\frac34$.
Note however that this reasoning works assuming you know what $|\psi\rangle$ is. Otherwise, the initial density matrix in the first case is $\frac{1}{2^{n-1}\left(2^n+1\right)}P_{\text{Sym}^2}\left(\mathbb{C}^{2^n}\right)$ as explained in this answer, with $P_{\text{Sym}^2}\left(\mathbb{C}^{2^n}\right)$ being the projector on the symmetric subspace of $\mathbb{C}^{2^n}$ with $2$ copies.
