Yes, it is true. More generally, the entries $1,2,3,0$ can be replaced by arbitrary numbers $a,b,c,d$, in which case the determinant of $M_n$ can be computed in terms of the four numbers $u = d-b$, $v = a-c$, $U = d+b$ and $V = a+c$ as follows:

If $n=4k$ for some positive integer $k$, then
$$
\det\left( M_{n} \right) =\dfrac{1}{4}
v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left(
2k-1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) .
$$

If $n=4k+2$ for some positive integer $k$, then
$$
\det\left( M_{n} \right) =-\dfrac{1}{4}%
v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left(
2k+1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) .
$$

If $n=4k+1$ for some positive integer $k$, then
$$
\det\left( M_{n} \right) =\dfrac{1}{2}
u^{n-3}\left( u^{2}\left( v+V\right) -\left( 2k\right) ^{2}v\left(
U^{2}-V^{2}\right) \right) .
$$

If $n=4k+3$ for some positive integer $k$, then
$$
\det\left( M_{n} \right) =\dfrac{1}{2}
vu^{n-3}\left( u^{2}+vV-\left( 2k+1\right) ^{2}\left( U^{2}-V^{2}\right)
\right) .
$$

As you can imagine, this is not very fun to prove. I have a writeup (*The 4-periodic spiral determinant*) in which I attempt at making the idea clear without going into all the details; in particular, annoying computations are relegated to SageMath and to the reader (and on some occasions to a combination of both). Even at that level of terseness, it is 24 pages long. I would normally hope that something nicer can be found, but with the complexity of the answer I am not too hopeful.

The proof starts out as suggested by @user44191 in one of the comments to the original post; thus the matrix is brought to a form where all entries are zero except for those in northwesternmost $4\times 4$-submatrix and on four sub-antidiagonals (namely, the $1$-st, the $3$-rd, the $5$-th and the $7$-th sub-antidiagonals) below the main antidiagonal. Then, I turn the matrix upside down, so that the sub-antidiagonals become the super-diagonals. I then perform Laplace expansion with respect to the last $4$ rows. All $4 \times 4$-minors from the last $4$ rows can be explicitly computed (only $\dbinom{7}{4}$ of them nonzero, and this can be further reduced by looking at the vanishing of the complementary minors), so it remains to compute the complementary $\left(n-4\right)\times\left(n-4\right)$-minors. For this, Jacobi's complementary minor theorem turns out to be of use, along with an explicit computation of the inverse of a certain power series.