They are not quasi-isomorphic: your $\check{\mathbf Z}(n)$ is concentrated in a single degree. As Denis points out in the comments, your definition of $\check{\mathbf Z}(n)$ is wrong because you should use the tensor product of sheaves with transfers. In addition, you need to apply Suslin's $\mathbf A^1$-invariantification construction $C_*$ to the result.

Voevodsky's motivic complex $\mathbf Z(n)_V$ is
$$\mathbf Z(n)_V = C_*(\mathbf G_m^{\otimes_{\mathrm{tr}}n})[-n],$$
where:

- $\mathbf G_m$ is regarded as a presheaf with transfers, i.e., a presheaf on Voevodsky's category $\mathrm{Cor}_k$ of smooth separated $k$-schemes and finite correspondences. The transfers are given by norms of invertible functions.
- $\otimes_{\mathrm{tr}}$ is the tensor product of sheaves with transfers, which is the Day convolution of the tensor product $X\otimes Y=X\times Y$ on $\mathrm{Cor}_k$.
- $C_*(F)(X)$ is a chain complex concentrated in nonnegative degrees with $C_n(F)(X)=F(X\times \mathbf A^n)$.

$\mathbf Z(n)_V$ is quasi-isomorphic to Bloch's $\mathbf Z(n)$ as a complex of Zariski sheaves on smooth $k$-schemes. This combines several deep results of Voevodsky, Suslin, and Friedlander. A more or less self-contained proof is in Mazza–Voevodsky-Weibel's Lecture notes on motivic cohomology.

On the other hand, $\mathbf Z(n)_V$ is also a complex of étale sheaves, and if $m$ is prime to the characteristic, then $\mathbf Z/m(n)_V$ is quasi-isomorphic to $\mu_m^{\otimes n}$ as a complex of étale sheaves. This is Theorem 10.3 in the above book. Perhaps this answers your last question.

**Edit:** Actually the tensor product in the above formula for $\mathbf Z(n)_V$ must be derived, so it's not very explicit. The "official" definition is
$$\mathbf Z(n)_V = C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n}))[-n],$$
where $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n})$ means the quotient of the sheaf $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n})$ sending $U$ to $\mathrm{Cor}_k(U,\mathbf G_m^{\times n})$ by the subsheaf generated by $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n-i-1}\times\{1\}\times\mathbf G_m^{\times i})$. The relation with the other formula comes from the fact that $C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge 1}))\to\mathbf G_m$ is a free resolution of $\mathbf G_m$ as a sheaf with transfers.