There is a purely geometrical explanation. First, let $n=2$, and consider $\mathrm{SL}(2,\mathbb{R})$.
Let $X : [0,1) \to \mathrm{SL}(2,\mathbb{R})$ be a smooth path, with $X(0) =E$, where $E$ is the identity.
$$X(t) = \left[\begin{array}{cc} a(t) & b(t) \\ c(t) & d(t) \end{array}\right]$$
where $\det[X(t)] = (ad-bc)(t) = 1$ for all $t$. The derivative $X'(t)$, as $t$ tends to zero, gives a tangent vector to $\mathrm{SL}(2,\mathbb{R})$ at $E$. We have
$$X'(t) = \left[\begin{array}{cc} a'(t) & b'(t) \\ c'(t) & d'(t) \end{array}\right]$$
Since $ad-bc \equiv 1$ we can differentiate to give
$$a'd+ad'-b'c-bc'\equiv 0$$
Since $X(0)=E$ we know that $a(0)=1$, $b(0)=0$, $c(0)=0$ and $d(0)=1$. Hence:
$$\begin{eqnarray*}
a'd+ad'-b'c-bc' &\equiv& 0 \\ \\
a'(0)\,d(0)+a(0)\,d'(0)-b'(0)\,c(0)-b(0)\,c'(0) &=& 0 \\ \\
a'(0)+d'(0) &=& 0
\end{eqnarray*}$$
This tells us that $X'(0)$ must be a traceless matrix.
General Dimension
Jacobi's Formula tells us that
$$\det(X)' = \mathrm{tr}\! \left[ \mathrm{adj}(X)X' \right] $$
Since $\det(X) \equiv 1$ we have $\det(X)'\equiv 0$. When $t=0$, $X=E$ and so $\mathrm{adj}(X) = E$. Hence
$$\mathrm{tr}\!\left[X'(0)\right]=0$$
Here the "the conditions that define" the Lie group are $a_{ij}=0, i<j$ differentiate that you obtain again $a_{ij}=0$ with no other constraints.
Generally, you can't have an invertibility constraint in a Lie algebra since it is a vector space.
Best Answer
The same way $(0,\frac12)$ can be a tangent vector to the unit circle at $(1,0)$ even though $(0,\frac12)$ is not on the unit circle.
One (loose and informal!) way to think about it is that an element of $\mathfrak{so}(3)$ is the difference between the matrix of an infinitesimal rotation and the identity matrix, but "scaled up by a factor of infinity" such that the entries of the matrix don't need to be infintesimals themselves.