The key to solving both problems is the use of the following two facts: 1) Any
closed subgroup of $T^n$ is the intersection of the kernels of characters of
$T^n$, i.e., continuous group homomorphisms $T^n \rightarrow S^1$. 2) Any
continuous homomorphism $T^n \rightarrow S^1$ is of the form $(\overline
x)\mapsto e^{2\pi i x\cdot m}$ for a unique $m\in\mathbb Z^n$. Hence, the
closure of $T_{\overline \nu}$ is the intersection of the kernels of the
characters corresponding to $m$ for which $\nu\cdot m\in\mathbb Z$. Picking a
basis $m_1,\dots,m_k$ of the group of such $m$ gives a surjection $T^n
\rightarrow T^k$ for which $T_{\overline\nu}$ is the kernel ($T_\nu$ is not
necessarily a torus as it might not be connected but it doesn't change
anything). By the above this map is just given by an $n\times k$ integer matrix
specified by $m_1,\dots,m_k$. The tangent map at the origin is then obtained by
regarding this matrix as a real matrix and thus the tangent space of
$T_{\overline \nu}$ is the null space of this matrix.

In particular this gives that the dimension of $T_{\overline \nu}$ is equal to
$\dim_{\mathbb Q}\langle1,\nu_1,\nu_2,\ldots,\nu_n\rangle -1$. This is off by one
from your guess if $1$ is in the span of of the $\nu_i$ but equal to it if it isn't.