I won't need to show a matrix this time. For a square $n$ by $n$ matrix with $n \geq 3,$ let $M$ be the symmetric matrix with all off-diagonal entries equal to $-1$ and all diagonal entries equal to $n-2.$ It is not difficult to show, using the eigenvalues, that all the principal minors up to size $n-1$ are nonnegative, so those submatrices are positive semidefinite. However, once again we get an eigenvalue of $-1$ with the eigenvector made up of all $1$'s. To get the submatrices positive definite, make the diagonal entries $n-2+ \delta$ with $0 < \delta < 1.$
Lost my nerve:
$$
\left( \begin{array}{rrrr}
2 & -1 & -1 & -1 \\
-1 & 2 & -1 & -1 \\
-1 & -1 & 2 & -1 \\
-1 & -1 & -1 & 2
\end{array}
\right)
$$
and
$$
\left( \begin{array}{rrrrr}
3 & -1 & -1 & -1 & -1 \\
-1 & 3 & -1 & -1 & -1 \\
-1 & -1 & 3 & -1 & -1 \\
-1 & -1 & -1 & 3 & -1 \\
-1 & -1 & -1 & -1 & 3
\end{array}
\right)
$$
Anyway, a $k$ by $k$ matrix consisting of all $1$'s has eigenvalues $0,0,0,\ldots, 0,k.$ If all the entries are $-1$ instead the eigenvalues are $0,0,0,\ldots, 0,-k.$ In order to put some number $w$ on the diagonal we need to add $(1 + w)I,$ where $I$ is the identity matrix. So the eigenvalues of the $k$ by $k$ matrix with all off-diagonal entries $-1$ and all diagonal entries $w$ are $1+w, \; 1+w, \; 1+w,\ldots, \; 1+w, \; 1 + w-k.$ This is how you quickly confirm the example above.
The positive definiteness (as you already pointed out) is a property of quadratic forms. However, there is a "natural" one-to-one correspondence between symmetric matrices and quadratic forms, so I really cannot see any reason why not to "decorate" symmetric matrices with positive definiteness (and other similar adjectives) just because it is in "reality" the form they define which actually has this property. I can see this one-to-one correspondence as one of the reasons why the symmetry should be implicitly assumed when talking about positive definite matrices.
One can of course devise a different name for this property, but why? In addition, positive definite matrix is a pretty standard term so if you continue reading on matrices I'm sure you will find it more and more often.
Some authors (not only on Math.SE) allow positive definite matrices to be nonsymmetric by saying that $M$ is such that $x^TMx>0$ for all nonzero $x$. In my opinion this adds more confusion than good (not only on Math.SE). Also note that (with a properly "fixed" inner product) such a definition would not even make sense in the complex case if the matrix was allowed to be non-Hermitian ($x^*Mx$ is real for all $x$ if and only if...).
Anyway, for real matrices, it of course makes sense to study nonsymmetric matrices giving a positive definite quadratic form through $x^TMx$ (which effectively means that the symmetric part is positive definite). However, I find denoting them as positive definite quite unlucky.
Best Answer
Your result is false: take $A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ then $A+ A^T=2I_2$ has his eigenvalues positive whereas eigenvalues of $A$ are $1 \pm i$ not even real...