I know that basis vector and basis of one-forms are related through
$$ \tilde{e}^\mu \cdot \vec{e}_\nu = \delta^\mu _\nu .\tag{1}$$
However, the metric has the property that allows to convert vectors to one-forms. So, can I try to say this: $$ \tilde{e}^\mu = g^{\mu \nu} \vec{e}_\nu , \tag{2}$$
if not explain me, please, maybe I don't clear with subscripts or can be some further.
Answer
First I'll state three quick preliminaries so we're both on the same page, and then I'll answer the question. In the following, I'm going to use tildes to distinguish one-forms and their components from vectors and their components; it is traditional to drop this extra notation and simply write $X_\mu = g_{\mu \nu}X^\nu$ as though the $X$'s on either side are the same object. However, since they are emphatically not the same object, I use the notation $\tilde X_\mu = g_{\mu \nu} X^\nu$ to make this explicit.
Preliminary #1: The Metric
The metric $\mathbf{g}$ is an object which linearly eats two vectors and spits out a scalar (e.g. a real number). The components of $\mathbf{g}$ in a particular basis are what you get when you feed the metric the basis vectors:
$$g_{\mu \nu} \equiv \mathbf{g}(\hat e_\mu,\hat e_\nu)$$
Therefore, you often see the action of $\mathbf{g}$ on two vectors $\mathbf{X}=X^\mu \hat e_\mu$ and $\mathbf{Y} = Y^\nu \hat e_\nu$ written like this:
$$\mathbf{g}(\mathbf X,\mathbf Y)=\mathbf{g}(X^\mu \hat e_\mu,Y^\nu \hat e_\nu) = X^\mu Y^\nu \mathbf{g}(\hat e_\mu,\hat e_\nu) = X^\mu Y^\nu g_{\mu \nu}$$
We can pull the components $X^\mu,Y^\nu$ out front because $\mathbf g$ is linear.
Preliminary #2: One-forms
A one-form, or covector, is an object which linearly eats a vector and spits out a scalar. Note that the following is a one-form:
$$\tilde{\mathbf X} := \mathbf{g}(\mathbf{X},\bullet)$$
From a vector $\mathbf X$ and the metric $\mathbf g$, we can construct a one-form $\tilde{\mathbf X}$ by plugging $\mathbf{X}$ into the first slot of $\mathbf g$ and leaving the second slot empty. We say that the one-form $\tilde{\mathbf X}$ is dual to the vector $\mathbf X$.
$\tilde{\mathbf X}$ then acts on some vector $\mathbf Y$ in the obvious way:
$$\tilde{\mathbf X}(\mathbf Y) = \mathbf{g}(\mathbf X,\mathbf Y)$$
In particular, if we feed $\tilde{\mathbf X}$ a basis vector $\hat e_\nu$, we get
$$\tilde{\mathbf X}(\hat e_\nu) = \mathbf{g}(\mathbf X,\hat e_\nu) = X^\mu \mathbf{g}(\hat e_\mu, \hat e_\nu) = X^\mu g_{\mu \nu}$$ If we define the one-form basis $\hat \epsilon^\mu$ to have the property $\hat \epsilon^\mu (\hat e_\nu) = \delta^\mu_\nu$, then we can expand $\tilde{\mathbf X}$ in its components $\tilde X_\mu$. Performing the same action,
$$\tilde{\mathbf X}(\hat e_\nu) = \tilde X_\mu \hat \epsilon^\mu(\hat e_\nu) = \tilde X_\mu \delta^\mu_\nu = \tilde X_\nu$$
Comparing this to what we got before, we see that $\tilde X_\nu = X^\mu g_{\mu\nu}$.
Note that even though we say one forms eat vectors and spit out scalars, we can also say that vectors eat one-forms and spit out scalars. We simply define the action of a vector $\mathbf X$ on a one-form $\tilde{\mathbf Y}$ to be $$\mathbf X(\tilde{\mathbf Y}) := \tilde{\mathbf Y}(\mathbf X)$$ This will be relevant in a moment.
Preliminary #3: The Inverse Metric
We've seen that we can use the metric to associate a vector $\mathbf X$ to a dual one-form $\tilde{\mathbf X}$; we can also go the other direction and associate a covector $\tilde{\mathbf Y}$ to a dual vector $\mathbf Y$. We do this by defining the so-called inverse metric $\tilde{\mathbf g}$, which is a map which eats two one-forms and spits out a scalar.
Essentially this is just a metric on the space of one-forms in exactly the same way as $\mathbf g$ is a metric on the space of vectors. However, we link them together by demanding that if the vectors $\mathbf X$ and $\mathbf Y$ have one-form duals $\tilde{\mathbf X}$ and $\tilde{\mathbf Y}$, then
$$\mathbf{g}(\mathbf X,\mathbf Y) = \tilde{\mathbf g}(\tilde{\mathbf X},\tilde{\mathbf Y})$$
It's a straightforward exercise to show that this means that the components of the dual metric $\tilde g^{\mu \nu}$ satisfy the following relationship to the components of the metric:
$$\tilde g^{\mu \nu} g_{\nu \rho} = \delta^\mu_\rho$$
This means that if we express them in matrix form, the $\tilde g^{\mu\nu}$'s are the matrix inverse of the $g_{\mu\nu}$'s - hence the name "inverse metric". We can now use this to associate $\tilde{\mathbf Y}$ with a vector:
$$\mathbf Y = \tilde{\mathbf g}(\tilde{\mathbf{Y}},\bullet)$$ It's straightforward to demonstrate (by feeding the basis one-form $\hat\epsilon^\mu$ to $\mathbf Y$ as defined above) that the components of $\mathbf Y$ are, as expected, given by $$Y^\mu = \tilde{g}^{\mu \nu}\tilde Y_\nu$$
Now your question can be answered. It's true that the metric can "convert" vectors into one-forms in an abstract sense. However, when we talk about "raising" and "lowering" indices like you are doing, we are converting the components of a vector to the components of the corresponding one-form.
The expression that you wrote ($\tilde g^{\mu \nu} \hat e_\nu$, in my notation) is simply a linear combination of vectors, and is therefore not a one-form as defined here. It is, however, what you get when you convert the basis one-form $\hat \epsilon^\mu$ into a vector, which I will now show.
Note that the one-form dual to the unit vector $\hat e_\mu$ is not the basis one-form $\hat \epsilon^\mu$; it is the one-form
$$\tilde{\boldsymbol \omega} := \mathbf{g}(\hat e_\mu, \bullet)$$
which has components
$$\tilde \omega_{\nu} \equiv \tilde{\boldsymbol\omega}(\hat e_\nu) = \mathbf{g}(\hat e_\mu, \hat e_\nu) = g_{\mu \nu}$$
To be concrete, the dual to the basis vector $\hat e_0$ is the one form $$\tilde{\boldsymbol \omega} = g_{0\nu}\hat \epsilon^\nu = g_{00}\hat \epsilon^0 + g_{01}\hat \epsilon^1 + g_{02}\hat \epsilon^2 + g_{03} \hat \epsilon^3$$
Using the inverse metric in precisely the same way, the dual to the basis one-form $\hat \epsilon^0$ is the vector
$$\boldsymbol \omega = \tilde g^{0\nu}\hat e_\nu = \tilde g^{00}\hat e_0 + \tilde g^{01} \hat e_1 + \tilde g^{02} \hat e_2 + \tilde g^{03} \hat e_3$$
To conclude, we do not have that $$\hat\epsilon^\mu = g^{\mu \nu} \hat e_\nu$$ but rather that $$\underbrace{\tilde{\mathbf g}(\hat \epsilon^\mu,\bullet)}_{\text{The vector *dual* to the basis one-form }\hat\epsilon^\mu} = g^{\mu \nu}\hat e_\nu$$
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