I just sum up the answers given above to refer to my question as answered:
My favorite argument is the one given by commenter here using Krein-Milmann theorem to prove that $C_0(K)$ has no predual space.
A good reference for such questions seems to be Topics in in Banach Space Theory by Kalton and Albiac. It is used here and here to prove that $C_0(K)$ and $L_1[0,1]$ have no predual space.
You can't really use that result as is, simply because the domain of $+$ is not $X$, but $X \times X$. You would really need to show that $\overline{X} \times \overline{X}$ is really the closure of $X \times X$. Of course, to do that, you'd need to nominate an appropriate norm (or a metric, at least) on $X \times X$.
I'm not convinced that it's worth doing. Instead, recall (or prove) that the sum of Cauchy sequences is Cauchy. Note that, if $(x_n), (y_n), (z_n)$ are Cauchy sequences with $(x_n)$ and $(y_n)$ equivalent, i.e. $\|x_n - y_n\| \to 0$, then $(x_n + z_n)$ and $(y_n + z_n)$ are equivalent (pretty obviously). This gives us a well-defined $+$ operation on the completion: $[x_n] + [y_n] := [x_n + y_n]$. You can do something similar with scalar multiplication.
Next comes the legwork. You have to show that $\overline{X}$ is a vector space with respect to the two operations. That means verifying all the axioms of a vector space, bar none. It's not a subspace of a known vector space, so you can't just show closure and non-emptiness.
You can use the theorem to extend the norm of $X$ to $\overline{X}$ (note that $\|\cdot\|$ is uniformly continuous, as $|\|x\| - \|y\|| \le \|x - y\|$). You then need to prove that this is a norm, using the axioms.
To conclude completeness, simply note that the metric on $\overline{X}$ agrees with the norm metric. You know that the completion is complete, hence $\overline{X}$ will therefore be a Banach space under the given norm.
I've brushed over a lot of detail! It's too much for me to write out, but none of it is tricky. In fact, most of it follows directly from $X$ being a vector space.
Best Answer
Sure. By Hahn-Banach theorem, we have that for any $x\in X$ it is $$\|x\|=\sup\{|\phi(x)|: \phi\in X^*, \|\phi\|=1\}.$$ If this is not known to you, I can add some details, but I recommend you do this as an exercise.
Now consider the mapping $u:X\to X^{**}$ that maps $x$ to the element $u_x:X^*\to\mathbb{C}$ ($X^{**}$ is the dual of the dual, so its elements are bounded linear maps from $X^*$ to $\mathbb{C})$ that is defined as follows: $u_x(\phi)=\phi(x)$ (this makes sense: $x\in X$ and $\phi\in X^*$, so $\phi(x)\in\mathbb{C}$).
Fix $x\in X$. We need to show that $u_x\in X^{**}$. It is immediate that $u_x$ is linear, as $u_x(\lambda_1\phi_1+\lambda_2\phi_2)=(\lambda_1\phi_1+\lambda_2\phi_2)(x)=\lambda_1\phi_1(x)+\lambda_2\phi_2(x)=\lambda_1u_x(\phi_1)+\lambda_2u_x(\phi_2)$. To see that $u_x$ is bounded on $X^{*}$, note that $|u_x(\phi)|=|\phi(x)|\leq\|\phi\|\|x\|$, which is of the form $c\|\phi\|$. So $u_x\in X^{**}$ with $\|u_x\|\leq\|x\|$.
Actually the 1st observation shows that $\|u_x\|=\|x\|$. Now it is immediate that $u:X\to X^{**}$ is linear. Since we observed that $\|u_x\|=\|x\|$, the map $u$ is an isometry. So set $\hat{X}=\overline{u(X)}$, the closure being taken in the norm of $X^{**}$. Do you see why $\hat{X}$ satisfies all those you mention?