Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,

Claim: Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even.

Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.

I don't know what's wrong with the following computation, but the answer is clearly no: if there were a cofinal functor from a finite simplicial set to $Idem$, then any $\infty$-category with finite colimits would have split idempotents, which is not the case (witness finite spaces).

Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,

Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.

Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,

Claim: Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even.

Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$, and it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.

Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,

Claim: Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even.

Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.