I am looking for a left invariant metric on $SL_n(\mathbb{R})$. If this is not possible, it would be acceptable to have a metric on $SL_n(\mathbb{R})/SO_n(\mathbb{R})$ or something like that. Is there such a thing?
Note: I am not looking for a Riemannian metric, just an ordinary metric.
The OP specifically asked for a(n ordinary) metric, not a Riemannian metric. While Misha and Paul have given good answers, I think that it's worth pointing out that, if one just takes an arbitrary left-invariant Riemannian metric $ds^2$ on a connected group $G$, there is no guarantee that the associated $dist$ can be computed explicitly. To do this, one would need to be able to integrate the geodesic equations explicitly enough to be able to construct the distance function; unless the Riemannian metric is quite special, this generally can't be done in any explicit way. (See for example, what one has to do to describe the free rotations of a rigid body that has 3 distinct moments of inertia, which is essentially computing the geodesics on $\text{SO}(3)$ with respect to a left-invariant metric that is not bi-invariant.)
An $\text{SL}(n,\mathbb{R})$-invariant metric on $\text{SL}(n,\mathbb{R})/\text{SO}(n)$: Since $M=\text{SL}(n,\mathbb{R})/\text{SO}(n)$ is an irreducible Riemannian symmetric space (with nonpositive sectional curvature), it has, up to constant multiples, only one $\text{SL}(n,\mathbb{R})$-invariant Riemannian metric, and the associated $dist$ in this case is not entirely trivial to write down: We can identify each element $m = A\cdot \text{SO}(n)$ with its associated positive definite, unimodular symmetric matrix $s = \sigma(m) = AA^T$, and the formula for $dist(s_1,s_2)$ is as follows: Write $$ \text{det}(ts_1-s_2) = (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n) $$ where each $\lambda_i$ is positive and they satisfy $\lambda_1\lambda_2\cdots\lambda_n=1$. Then, up to a constant multiple, one has $$ dist(s_1,s_2) = \left(\sum_{i=1}^n (\log\lambda_i)^2\right)^{1/2}. $$ Obviously, writing this out as a function of the entries of $s_1$ and $s_2$ would not be easy. (Of course, this is just one example of the sort of metric that Paul gave; its distinguishing characteristic is that it is the $dist$ of a Riemannian metric, which is not true for Paul's specific example.)
A left-invariant metric on $\text{SL}(n,\mathbb{R})$: Once an invariant metric on $M$ has been defined, one can use it to define a metric on $\text{SL}(n,\mathbb{R})$ itself: First, suppose that $n$ is odd, so that $\text{SL}(n,\mathbb{R})$ acts effectively on $M$. Let $\delta:M\times M\to \mathbb{R}$ be an invariant metric and let $(s_1,\ldots,s_k)\in M\times M\times \cdots \times M$ ($k$ times) be a $k$-tuple of symmetric matrices with the property that the simultaneous stabilizer of all of the $s_i$ in $\text{SL}(n,\mathbb{R})$ is the identity matrix. (I guess $k=3$ suffices to find such a $k$-tuple; $k=2$ does not.) Set $$ dist(A,B) = \delta(As_1,Bs_1) + \delta(As_2,Bs_2) + \cdots + \delta(As_k,Bs_k). $$ This defines a left-invariant metric on $\text{SL}(n,\mathbb{R})$. Note, however, that this is not derived from a Riemannian metric. When $n$ is even, $-I_n$ lies in $\text{SL}(n,\mathbb{R})$ and it acts trivially on $M$, so the above construction won't work. However, when $n$ is even, just take the metric induced on $\text{SL}(n,\mathbb{R})$ by its natural embedding into $\text{SL}(n{+}1,\mathbb{R})$, and that will do the job.
To get a Riemannian metric on $\text{SL}(n,\mathbb{R})$ whose $dist$ is computable, one should probably take the left-invariant metric $ds^2 = \text{tr}\bigl((g^{-1}dg)^Tg^{-1}dg\bigr)$. The reason is that this metric is both left-invariant and invariant under right action by $\text{SO}(n)$, and the extra symmetries make the geodesic flow explicitly integrable. (There is, of course, no bi-invariant Riemannian metric on $\text{SL}(n,\mathbb{R})$.)
A little calculation shows that the $ds^2$-geodesic starting at $I_n$ with velocity $v\in{\frak{sl}}(n,\mathbb{R})$ is given by the formula $$ \gamma_v(t) = e^{v^Tt}e^{(v-v^T)t}, $$ where $v^T$ represents the transpose of $v$. In particular, one has the 'formula', for $A\in\mathrm{SL}(n,\mathbb{R})$, $$ dist(I_n,A) = \min\bigl\{\bigl(\text{tr}(v^Tv)\bigr)^{1/2}\ |\ e^{v^T}e^{(v-v^T)} = A\bigr\}. $$ Unfortunately, computing this $dist$ more explicitly is a challenge. By left-invariance, of course, one has $dist(A,B) = dist(I_n,A^{-1}B)$, so this determines the metric completely.
Computing $dist$ explicitly is nontrivial even in the case $n=2$. In this case, one has the fortunate circumstance that, unless $\det(v)>0$, there are no conjugate points along the geodesic $\gamma_v$, and, when $\det(v)>0$, the first conjugate point is at $t = \pi/\sqrt{\det(v)}$. However, describing the exact cut locus of $I_n$ in ${\frak{sl}}(n,\mathbb{R})$ with respect to $ds^2$ does not seem to be trivial. This does not seem to be a particularly good way to construct a left-invariant metric on $\text{SL}(2,\mathbb{R})$. Nevertheless, it's not hopeless. If I have time, I'll add a little note to this describing what one can say.
Just to clear the air: $SL(n,R)$, as any Lie group $G$, admits a left-invariant Riemannian metric $ds^2$. This metric is not unique, but it is determined by $ds^2$ at $T_eG$, i.e., on the Lie algebra of $G$. If $G$ is connected (and $SL(n,R)$, of course, is), then $ds^2$ determines (canonically) a distance function $dist$ on $G$. Since $ds^2$ was left-invariant, so is $dist$. This answers OP's question.
If one were do deal with disconnected Lie groups, then one would have to make some further choices to define a left-invariant distance function on $G$. Here is one construction modelled on what one does in geometric group theory. Let $S\subset G$ be a smallest subset so that $G_0 \cup S$ generates $G$. Here $G_0\subset G$ is the component of identity. Then, define a "Cayley graph", where we connect elements $g, gs$ by an edge of unit length, for all $g\in G, s\in S$. Now, every two points of $G$ are connected by a path which is a concatenation of smooth paths in components of $G$ and edges of the graph. Finally, take the length-metric on $G$ defined using such paths, by minimizing lengths of paths connecting two points $x,y\in G$. By construction, this metric will be a left-invariant.
There is a (left) $G$-invariant metric on $G/K$ for $G$ reductive, $K$ maximal compact, admitting a Cartan-type decomposition $G=KA^+K$ with $A^+$ the connected component of the identity in a maximal real-split torus (e.g., positive real diagonal matrices for $G=SL_n(\mathbb R)$. For $G=SL_n(\mathbb R)$, let $m(kak')$ be the max of the diagonal entries of $a\in A^+$ and of their inverses. Then $d(gK,hK)=\log m(h^{-1}g)$ is a metric on $G/K$.
The triangle inequality follows from the submultiplicativity of the operator norm $|\cdot|$, and noting that $m(g)=\max |g|,|g^{-1}|$.
